Chapter 29 : The plane - Rd Sharma Solutions for Class 12-science Maths CBSE

Chapter 29 - The plane Excercise Ex. 29.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Chapter 29 - The plane Excercise Ex. 29.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Chapter 29 - The plane Excercise Ex. 29.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Find the vector and Cartesian equations of the plane which passes through the point (5, 2, -4) and perpendicular to the line with direction ratios 2, 3, -1.

Solution 22

begin mathsize 12px style table attributes columnalign left end attributes row cell text Vector   equation   of   the   plane : end text end cell row cell text Given   that   the   required   plane   passes   through   the end text end cell row cell text point   end text left parenthesis text 5 , 2 , end text minus 4 right parenthesis text   having   the   position   vector end text end cell row cell stack text a end text with rightwards arrow on top equals 5 straight i with hat on top plus 2 straight j with hat on top minus 4 straight k with hat on top end cell row cell Also text   given   that   the   required   plane   is   perpendicular end text end cell row cell text to   the   line   with   direction   ratios   2 ,  3   and   end text minus 1. end cell row cell Thus text   the   vector   equation   of   the   normal   vector   to   the end text end cell row cell text plane   is   end text stack text n end text with rightwards arrow on top equals 2 straight i with hat on top plus 3 straight j with hat on top minus straight k with hat on top. end cell row cell text We   know   that   the   vector   equation   of   the   plane   passing end text end cell row cell text through   a   point   having   position   vector   end text stack text a end text with rightwards arrow on top text   and   normal   to end text end cell row cell text vector   end text stack text n end text with rightwards arrow on top text   is   given   by   end text left parenthesis stack text r end text with rightwards arrow on top minus stack text a end text with rightwards arrow on top right parenthesis times stack text n end text with rightwards arrow on top equals 0 text   or ,  end text stack text r end text with rightwards arrow on top times stack text n end text with rightwards arrow on top equals stack text a end text with rightwards arrow on top times stack text n end text with rightwards arrow on top. end cell row cell Thus text   the   required   equation   of   the   required   plane   is end text end cell row cell stack text r end text with rightwards arrow on top times left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top minus straight k with hat on top right parenthesis equals left parenthesis 5 straight i with hat on top plus 2 straight j with hat on top minus 4 straight k with hat on top right parenthesis times left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top minus straight k with hat on top right parenthesis end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top minus straight k with hat on top right parenthesis equals 10 plus 6 plus 4 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top minus straight k with hat on top right parenthesis equals 20 end cell end table end style

 

begin mathsize 12px style table attributes columnalign left end attributes row cell text The   Cartesian   equation   of   the   plane   is   end text end cell row cell stack text r end text with rightwards arrow on top times left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top minus straight k with hat on top right parenthesis equals 20 end cell row cell rightwards double arrow left parenthesis straight x straight i with hat on top plus straight y straight j with hat on top plus straight z straight k with hat on top right parenthesis times left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top minus straight k with hat on top right parenthesis equals 2 end cell row cell rightwards double arrow 2 straight x plus 3 straight y minus straight z equals 20 end cell end table end style

Question 23

If O be the origin and the coordinates of P be (1, 2, -3), then find the equation of the plane passing through P and perpendicular to OP.

Solution 23

begin mathsize 12px style table attributes columnalign left end attributes row cell text Consider   the   point    P end text left parenthesis text 1 , 2 , end text minus 3 right parenthesis. end cell row cell text Thus   the   position   vector   of   the   point   P   is end text end cell row cell stack text a end text with rightwards arrow on top equals straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top end cell row cell Direction text   ratios   of   the   line   OP ,  where   O   is   the   end text end cell row cell text origin ,  are   1 , 2   and   end text minus 3 end cell row cell Thus text   the   vector   equation   of   the   normal   vector ,  OP ,  to   the end text end cell row cell text plane   is   end text stack text n end text with rightwards arrow on top equals straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top. end cell row cell text We   know   that   the   vector   equation   of   the   plane   passing end text end cell row cell text through   a   point   having   position   vector   end text stack text a end text with rightwards arrow on top text   and   normal   to end text end cell row cell text vector   end text stack text n end text with rightwards arrow on top text   is   given   by   end text left parenthesis stack text r end text with rightwards arrow on top minus stack text a end text with rightwards arrow on top right parenthesis times stack text n end text with rightwards arrow on top equals 0 text   or ,  end text stack text r end text with rightwards arrow on top times stack text n end text with rightwards arrow on top equals stack text a end text with rightwards arrow on top times stack text n end text with rightwards arrow on top. end cell row cell Thus text   the   required   equation   of   the   required   plane   is end text end cell row cell stack text r end text with rightwards arrow on top times left parenthesis straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis equals left parenthesis straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis times left parenthesis straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times left parenthesis straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis equals 1 plus 4 plus 9 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times left parenthesis straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis equals 14 end cell row cell rightwards double arrow left parenthesis straight x straight i with hat on top plus straight y straight j with hat on top plus straight z straight k with hat on top right parenthesis times left parenthesis straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis equals 14 end cell row cell rightwards double arrow straight x plus 2 straight y minus 3 straight z equals 14 end cell end table end style

Question 24

If O is the origin and the coordinates of A are (a, b, c). Find the direction cosines of OA and the equation of the plane through A at right angles to OA.

Solution 24

  

Chapter 29 - The plane Excercise Ex. 29.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   the   vector   equation   of   the   plane   which   end text end cell row cell text is   at   a   distance   of   end text fraction numerator text 6 end text over denominator square root of text 29 end text end root end fraction text   from   the   origin   and   its   end text end cell row cell text normal   vector   from   the   origin   is   end text 2 straight i with hat on top minus 3 straight j with hat on top plus 4 straight k with hat on top. text   end text end cell row cell text Also ,  find   its   cartesian   form. end text end cell end table end style

Solution 10

begin mathsize 12px style table attributes columnalign left end attributes row cell We text   know   that   the   vector   equation   of   a   plane   at   a   distance end text end cell row cell text ' p '  from   the   origin   and   normal   to   the   unit   vector   end text stack text n end text with hat on top text   is   end text stack text r end text with rightwards arrow on top times stack text n end text with hat on top equals straight p end cell row cell Vector text   normal   to   the   plane   is   end text stack text n end text with rightwards arrow on top equals 2 straight i with hat on top minus 3 straight j with hat on top plus 4 straight k with hat on top end cell row cell The text    end text unit text   vector   normal   to   the   plane   is end text end cell row cell stack text n end text with hat on top equals fraction numerator 2 over denominator square root of 2 squared plus left parenthesis negative 3 right parenthesis squared plus 4 squared end root end fraction straight i with hat on top minus fraction numerator 3 over denominator square root of 2 squared plus left parenthesis negative 3 right parenthesis squared plus 4 squared end root end fraction straight j with hat on top plus fraction numerator 4 over denominator square root of 2 squared plus left parenthesis negative 3 right parenthesis squared plus 4 squared end root end fraction straight k with hat on top end cell row cell rightwards double arrow stack text n end text with hat on top equals fraction numerator 2 over denominator square root of 4 plus 9 plus 16 end root end fraction straight i with hat on top minus fraction numerator 3 over denominator square root of 4 plus 9 plus 16 end root end fraction straight j with hat on top plus fraction numerator 4 over denominator square root of 4 plus 9 plus 16 end root end fraction straight k with hat on top end cell row cell rightwards double arrow stack text n end text with hat on top equals fraction numerator 2 over denominator square root of 29 end fraction straight i with hat on top minus fraction numerator 3 over denominator square root of 29 end fraction straight j with hat on top plus fraction numerator 4 over denominator square root of 29 end fraction straight k with hat on top end cell row cell Here comma text   given   that   p = end text fraction numerator text 6 end text over denominator square root of text 29 end text end root end fraction end cell row cell Thus comma text   the   vector   equation   of   the   plane   is end text end cell row cell stack text r end text with rightwards arrow on top times open parentheses fraction numerator 2 over denominator square root of 29 end fraction straight i with hat on top minus fraction numerator 3 over denominator square root of 29 end fraction straight j with hat on top plus fraction numerator 4 over denominator square root of 29 end fraction straight k with hat on top close parentheses equals fraction numerator text 6 end text over denominator square root of text 29 end text end root end fraction end cell row cell The text   Cartesian   equation   of   the   plane   is end text end cell row cell open parentheses straight x straight i with hat on top plus straight y straight j with hat on top plus straight z straight k with hat on top close parentheses times open parentheses fraction numerator 2 over denominator square root of 29 end fraction straight i with hat on top minus fraction numerator 3 over denominator square root of 29 end fraction straight j with hat on top plus fraction numerator 4 over denominator square root of 29 end fraction straight k with hat on top close parentheses equals fraction numerator text 6 end text over denominator square root of text 29 end text end root end fraction end cell row cell rightwards double arrow open parentheses fraction numerator 2 straight x over denominator square root of 29 end fraction minus fraction numerator 3 straight y over denominator square root of 29 end fraction plus fraction numerator 4 straight z over denominator square root of 29 end fraction close parentheses equals fraction numerator text 6 end text over denominator square root of text 29 end text end root end fraction end cell row cell rightwards double arrow open parentheses fraction numerator 2 straight x minus 3 straight y plus 4 straight z over denominator square root of 29 end fraction close parentheses equals fraction numerator text 6 end text over denominator square root of text 29 end text end root end fraction end cell row cell rightwards double arrow 2 straight x minus 3 straight y plus 4 straight z equals text 6 end text end cell end table end style

Question 11

Find the distance of the plane 2x - 3y + 4z - 6 = 0 from the origin.

Solution 11

begin mathsize 12px style table attributes columnalign left end attributes row cell The text   Cartesian   equation   of   the   given   plane   is end text end cell row cell text 2 x end text minus 3 straight y plus 4 straight z minus 6 equals 0. end cell row cell The text   above   equation   can   be   rewritten   as end text end cell row cell text 2 x end text minus 3 straight y plus 4 straight z equals 6 end cell row cell Therefore comma text   end text the text   vector   equation   of   the   plane   is end text end cell row cell open parentheses straight x straight i with hat on top plus straight y straight j with hat on top plus straight z straight k with hat on top close parentheses times open parentheses 2 straight i with hat on top minus 3 straight j with hat on top plus 4 straight k with hat on top close parentheses equals 6 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times open parentheses 2 straight i with hat on top minus 3 straight j with hat on top plus 4 straight k with hat on top close parentheses equals 6.... left parenthesis 1 right parenthesis end cell row cell We text   know   that   the   vector   equation   of   a   plane   at   a   distance end text end cell row cell text ' p '  from   the   origin   and   normal   to   unit   vector   end text stack text n end text with hat on top text   is   end text stack text r end text with rightwards arrow on top times stack text n end text with hat on top equals straight p end cell row blank row cell We text   have ,  end text stack text n end text with rightwards arrow on top equals 2 straight i with hat on top minus 3 straight j with hat on top plus 4 straight k with hat on top. end cell row cell Thus text   end text vertical line stack text n end text with rightwards arrow on top vertical line equals square root of 2 squared plus left parenthesis negative 3 right parenthesis squared plus 4 squared end root equals square root of 29 end cell row blank row cell Dividing text   the   equation  ( 1 )  by   end text vertical line stack text n end text with rightwards arrow on top vertical line equals square root of text 29 end text end root comma text   we   have , end text end cell row cell stack text r end text with rightwards arrow on top times open parentheses fraction numerator 2 straight i with hat on top minus 3 straight j with hat on top plus 4 straight k with hat on top over denominator square root of 29 end fraction close parentheses equals fraction numerator 6 over denominator square root of 29 end fraction end cell row cell Hence text   the   normal   form   of   the   equation   of   the   plane   is end text end cell row cell stack text r end text with rightwards arrow on top times open parentheses fraction numerator 2 over denominator square root of 29 end fraction straight i with hat on top minus fraction numerator 3 over denominator square root of 29 end fraction straight j with hat on top plus fraction numerator 4 over denominator square root of 29 end fraction straight k with hat on top close parentheses equals fraction numerator 6 over denominator square root of 29 end fraction end cell row cell text Hence   the   perpendicular   distance   of   the   end text end cell row cell text origin   from   the   plane   is   p = end text fraction numerator text 6 end text over denominator square root of text 29 end text end root end fraction. end cell end table end style

Chapter 29 - The plane Excercise Ex. 29.5

Question 1

Solution 1

Question 2

Find the vector equation of the plane passing through the points P(2, 5, -3), Q(-2, -3, 5) and R(5, 3, -3).

Solution 2

begin mathsize 12px style table attributes columnalign left end attributes row cell Let text   P end text left parenthesis text 2 , 5 , end text minus text 3 end text right parenthesis comma text   Q end text left parenthesis negative 2 comma negative 3 comma 5 right parenthesis text   and   R end text left parenthesis text 5 , 3 , end text minus 3 right parenthesis text   be   the   three end text end cell row cell text points   on   a   plane   having   position   vectors   end text stack text p end text with rightwards arrow on top text ,  end text stack text q end text with rightwards arrow on top text   and   end text stack text s end text with rightwards arrow on top end cell row cell text respectively .  Then   the   vectors   end text stack text PQ end text with rightwards arrow on top text   and   end text stack text PR end text with rightwards arrow on top text   are   in   the   same   plane. end text end cell row cell text Therefore ,  end text stack text PQ end text with rightwards arrow on top cross times stack text PR end text with rightwards arrow on top text   is   a   vector   perpendicular   to   the   plane. end text end cell row cell text Let   end text stack text n end text with rightwards arrow on top text  =  end text stack text PQ end text with rightwards arrow on top cross times stack text PR end text with rightwards arrow on top end cell row blank row cell PQ with rightwards arrow on top equals left parenthesis negative 2 minus 2 right parenthesis straight i with hat on top plus left parenthesis negative 3 minus 5 right parenthesis straight j with hat on top plus left parenthesis 5 minus left parenthesis negative 3 right parenthesis right parenthesis straight k with hat on top end cell row cell rightwards double arrow PQ with rightwards arrow on top equals negative 4 straight i with hat on top minus 8 straight j with hat on top plus 8 straight k with hat on top end cell row cell Similarly comma end cell row cell PR with rightwards arrow on top equals left parenthesis 5 minus 2 right parenthesis straight i with hat on top plus left parenthesis 3 minus 5 right parenthesis straight j with hat on top plus left parenthesis negative 3 minus left parenthesis negative 3 right parenthesis right parenthesis straight k with hat on top end cell row cell rightwards double arrow PR with rightwards arrow on top equals 3 straight i with hat on top minus 2 straight j with hat on top plus 0 straight k with hat on top end cell row cell Thus text   end text end cell row cell stack text n end text with rightwards arrow on top equals stack text PQ end text with rightwards arrow on top cross times stack text PR end text with rightwards arrow on top end cell row cell text     end text equals open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row cell negative 4 end cell cell negative 8 end cell 8 row 3 cell negative 2 end cell 0 end table close vertical bar end cell row cell text     end text equals 16 straight i with hat on top plus 24 straight j with hat on top plus 32 straight k with hat on top end cell row cell The text   plane   passes   through   the   point   P   with   end text end cell row cell text position   vector   end text stack text p end text with rightwards arrow on top equals 2 straight i with hat on top plus 5 straight j with hat on top minus 3 straight k with hat on top end cell row cell text Thus ,  its   vector   equation   is end text end cell row cell left curly bracket stack text r end text with rightwards arrow on top minus left parenthesis 2 straight i with hat on top plus 5 straight j with hat on top minus 3 straight k with hat on top right parenthesis right curly bracket times left parenthesis 16 straight i with hat on top plus 24 straight j with hat on top plus 32 straight k with hat on top right parenthesis equals 0 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times left parenthesis 16 straight i with hat on top plus 24 straight j with hat on top plus 32 straight k with hat on top right parenthesis minus left parenthesis 32 plus 120 minus 96 right parenthesis equals 0 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times left parenthesis 16 straight i with hat on top plus 24 straight j with hat on top plus 32 straight k with hat on top right parenthesis minus 56 equals 0 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times left parenthesis 16 straight i with hat on top plus 24 straight j with hat on top plus 32 straight k with hat on top right parenthesis equals 56 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top plus 4 straight k with hat on top right parenthesis equals 7 end cell end table end style

Question 3

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   the   vector   equation   of   the   plane   passing   end text end cell row cell text through   the   points   end text straight A left parenthesis straight a text , 0 , 0 end text right parenthesis comma straight B left parenthesis 0 comma straight b comma 0 right parenthesis text   and   end text straight C left parenthesis 0 comma 0 comma straight c right parenthesis. text   end text end cell row cell text Reduce   it   to   normal   form .  end text end cell row cell text If   plane   end text ABC text   is   at   a   distance   end text straight p text   from   the   origin ,  end text end cell row cell text prove   that   end text fraction numerator text 1 end text over denominator straight p to the power of text 2 end text end exponent end fraction equals fraction numerator text 1 end text over denominator straight a to the power of text 2 end text end exponent end fraction plus fraction numerator text 1 end text over denominator straight b to the power of text 2 end text end exponent end fraction plus fraction numerator text 1 end text over denominator straight c to the power of text 2 end text end exponent end fraction. end cell end table end style

Solution 3

begin mathsize 12px style table attributes columnalign left end attributes row cell Let text   A end text left parenthesis text a , 0 , end text 0 right parenthesis comma text   B end text left parenthesis 0 comma straight b comma 0 right parenthesis text   and   C end text left parenthesis 0 comma 0 comma straight c right parenthesis text   be   three end text end cell row cell text points   on   a   plane   having   their   position   vectors   end text stack text a end text with rightwards arrow on top text ,  end text stack text b end text with rightwards arrow on top text   and   end text stack text c end text with rightwards arrow on top end cell row cell text respectively .  Then   vectors   end text stack text AB end text with rightwards arrow on top text   and   end text stack text AC end text with rightwards arrow on top text   are   in   the   same   plane. end text end cell row cell text Therefore ,  end text stack text AB end text with rightwards arrow on top cross times stack text AC end text with rightwards arrow on top text   is   a   vector   perpendicular   to   the   plane. end text end cell row cell text Let   end text stack text n end text with rightwards arrow on top text = end text stack text AB end text with rightwards arrow on top cross times stack text AC end text with rightwards arrow on top end cell row blank row cell stack text AB end text with rightwards arrow on top equals left parenthesis 0 minus straight a right parenthesis straight i with hat on top plus left parenthesis straight b minus 0 right parenthesis straight j with hat on top plus left parenthesis 0 minus 0 right parenthesis straight k with hat on top end cell row cell rightwards double arrow stack text AB end text with rightwards arrow on top equals negative straight a straight i with hat on top plus straight b straight j with hat on top plus 0 straight k with hat on top end cell row cell Similarly comma end cell row cell stack text AC end text with rightwards arrow on top equals left parenthesis 0 minus straight a right parenthesis straight i with hat on top plus left parenthesis 0 minus 0 right parenthesis straight j with hat on top plus left parenthesis straight c minus 0 right parenthesis straight k with hat on top end cell row cell rightwards double arrow stack text AC end text with rightwards arrow on top equals negative straight a straight i with hat on top plus 0 straight j with hat on top plus straight c straight k with hat on top end cell row cell Thus text   end text end cell row cell stack text n end text with rightwards arrow on top equals stack text AB end text with rightwards arrow on top cross times stack text AC end text with rightwards arrow on top end cell row cell text     end text equals vertical line table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row cell negative straight a end cell straight b 0 row cell negative straight a end cell 0 straight c end table vertical line end cell row cell stack text n end text with rightwards arrow on top equals bc straight i with hat on top plus ac straight j with hat on top plus ab straight k with hat on top end cell row cell rightwards double arrow straight n with hat on top equals fraction numerator bc straight i with hat on top plus ac straight j with hat on top plus ab straight k with hat on top over denominator square root of straight b squared straight c squared plus straight a squared straight c squared plus straight a squared straight b squared end root end fraction end cell row cell The text   plane   passes   through   the   point   P   with   end text end cell row cell text position   vector   end text stack text a end text with rightwards arrow on top equals straight a straight i with hat on top plus 0 straight j with hat on top plus 0 straight k with hat on top end cell end table end style

 

begin mathsize 12px style table attributes columnalign left end attributes row cell text Thus ,  the   vector   equation   in   the   normal   form   is end text end cell row cell open curly brackets stack text r end text with rightwards arrow on top minus open parentheses left parenthesis straight a straight i with hat on top plus 0 straight j with hat on top plus 0 straight k with hat on top close parentheses close curly brackets times open parentheses fraction numerator bc straight i with hat on top plus ac straight j with hat on top plus ab straight k with hat on top over denominator square root of straight b squared straight c squared plus straight a squared straight c squared plus straight a squared straight b squared end root end fraction close parentheses equals 0 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times fraction numerator left parenthesis bc straight i with hat on top plus ac straight j with hat on top plus ab straight k with hat on top right parenthesis over denominator square root of straight b squared straight c squared plus straight a squared straight c squared plus straight a squared straight b squared end root end fraction equals fraction numerator abc over denominator square root of straight b squared straight c squared plus straight a squared straight c squared plus straight a squared straight b squared end root end fraction end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times fraction numerator open parentheses bc straight i with hat on top plus ac straight j with hat on top plus ab straight k with hat on top close parentheses over denominator square root of straight b squared straight c squared plus straight a squared straight c squared plus straight a squared straight b squared end root end fraction equals fraction numerator 1 over denominator square root of fraction numerator straight b squared straight c squared plus straight a squared straight c squared plus straight a squared straight b squared over denominator straight a squared straight b squared straight c squared end fraction end root end fraction end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times fraction numerator left parenthesis bc straight i with hat on top plus ac straight j with hat on top plus ab straight k with hat on top right parenthesis over denominator square root of straight b squared straight c squared plus straight a squared straight c squared plus straight a squared straight b squared end root end fraction equals fraction numerator 1 over denominator square root of 1 over straight a squared plus 1 over straight b squared plus 1 over straight c squared end root end fraction... left parenthesis 1 right parenthesis end cell row cell The text   vector   equation   of   a   plane   normal   to   the   unit   vector end text end cell row cell stack text n end text with hat on top text   and   at   a   distance  ' d '  from   the   origin   is   end text stack text r end text with rightwards arrow on top times stack text n   end text with hat on top text =  d end text... text ( 2 ) end text end cell row cell Given text   that   the   plane   is   at   a   distance  ' p '  from   the end text end cell row cell text origin. end text end cell row cell text Comparing   equations  ( 1 )  and  ( 2 ),  we   have , end text end cell row cell text d  =  p  =  end text fraction numerator 1 over denominator square root of 1 over straight a squared plus 1 over straight b squared plus 1 over straight c squared end root end fraction end cell row cell rightwards double arrow 1 over straight p squared equals 1 over straight a squared plus 1 over straight b squared plus 1 over straight c squared end cell end table end style

Question 4

Find the vector equation of the plane passing through the points (1, 1, -1), (6, 4, -5) and (-4, -2, 3).

Solution 4

begin mathsize 12px style table attributes columnalign left end attributes row cell Let text   P end text left parenthesis text 1 , 1 , end text minus 1 right parenthesis comma text   Q end text left parenthesis 6 comma 4 comma negative 5 right parenthesis text   and   R end text left parenthesis negative text 4 , end text minus text 2 , end text 3 right parenthesis text   be   three end text end cell row cell text points   on   a   plane   having   position   vectors   end text stack text p end text with rightwards arrow on top text ,  end text stack text q end text with rightwards arrow on top text   and   end text stack text s end text with rightwards arrow on top end cell row cell text respectively .  Then   the   vectors   end text stack text PQ end text with rightwards arrow on top text   and   end text stack text PR end text with rightwards arrow on top text   are   in   the   same   plane. end text end cell row cell text Therefore ,  end text stack text PQ end text with rightwards arrow on top cross times stack text PR end text with rightwards arrow on top text   is   a   vector   perpendicular   to   the   plane. end text end cell row cell text Let   end text stack text n end text with rightwards arrow on top text  =  end text stack text PQ end text with rightwards arrow on top cross times stack text PR end text with rightwards arrow on top end cell row blank row cell PQ with rightwards arrow on top equals left parenthesis 6 minus 1 right parenthesis straight i with hat on top plus left parenthesis 4 minus 1 right parenthesis straight j with hat on top plus left parenthesis negative 5 minus left parenthesis negative 1 right parenthesis right parenthesis straight k with hat on top end cell row cell rightwards double arrow PQ with rightwards arrow on top equals 5 straight i with hat on top plus 3 straight j with hat on top minus 4 straight k with hat on top end cell row cell Similarly comma end cell row cell PR with rightwards arrow on top equals left parenthesis negative 4 minus 1 right parenthesis straight i with hat on top plus left parenthesis negative 2 minus 1 right parenthesis straight j with hat on top plus left parenthesis 3 minus left parenthesis negative 1 right parenthesis right parenthesis straight k with hat on top end cell row cell rightwards double arrow PR with rightwards arrow on top equals negative 5 straight i with hat on top minus 3 straight j with hat on top plus 4 straight k with hat on top end cell row cell Thus text   end text end cell row cell Here comma text   end text stack text PQ end text with rightwards arrow on top equals negative stack text PR end text with rightwards arrow on top text   end text end cell row cell Therefore comma text   the   given   points   are   collinear .  end text end cell row cell text Thus ,  end text stack text n end text with rightwards arrow on top equals straight a straight i with hat on top plus straight b straight j with hat on top plus straight c straight k with hat on top text   where ,  5 a + 3 b end text minus 4 straight c equals 0 end cell row cell The text   plane   passes   through   the   point   P   with   end text end cell row cell text position   vector   end text stack text p end text with rightwards arrow on top equals straight i with hat on top plus straight j with hat on top minus straight k with hat on top end cell row cell text Thus ,  its   vector   equation   is end text end cell row cell left curly bracket stack text r end text with rightwards arrow on top minus left parenthesis straight i with hat on top plus straight j with hat on top minus straight k with hat on top right parenthesis right curly bracket times left parenthesis straight a straight i with hat on top plus straight b straight j with hat on top plus straight c straight k with hat on top right parenthesis equals 0 comma text   where ,  5 a + 3 b end text minus 4 straight c equals 0 end cell end table end style

Question 5

Solution 5

Chapter 29 - The plane Excercise Ex. 29.6

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Find the angle between the planes:

2x + y - 2z = 5 and 3x - 6y - 2z = 7

Solution 8

begin mathsize 12px style table attributes columnalign left end attributes row cell We text   know   that   the   angle   between   the   planes end text end cell row cell text a end text subscript text 1 end text end subscript text x  +  b end text subscript text 1 end text end subscript text y  +  c end text subscript text 1 end text end subscript text z  +  d end text subscript text 1 end text end subscript text  =  0   and   a end text subscript text 2 end text end subscript text x  +  b end text subscript text 2 end text end subscript text y  +  c end text subscript text 2 end text end subscript text z  +  d end text subscript text 2 end text end subscript text  =  0   is   given   by end text end cell row cell text cos end text straight theta equals fraction numerator straight a subscript 1 straight a subscript 2 plus straight b subscript 1 straight b subscript 2 plus straight c subscript 1 straight c subscript 2 over denominator square root of straight a subscript 1 squared plus straight b subscript 1 squared plus straight c subscript 1 squared end root times square root of straight a subscript 2 squared plus straight b subscript 2 squared plus straight c subscript 2 squared end root end fraction end cell row cell Therefore comma text   the   angle   between   2 x + y end text minus 2 straight z equals 5 text   and   3 x end text minus text 6 y end text minus text 2 z = 7 end text end cell row cell text cos end text straight theta equals fraction numerator 2 cross times 3 plus 1 cross times left parenthesis negative 6 right parenthesis plus left parenthesis negative 2 right parenthesis cross times left parenthesis negative 2 right parenthesis over denominator square root of 2 squared plus 1 squared plus left parenthesis negative 2 right parenthesis squared end root times square root of 3 squared plus left parenthesis negative 6 right parenthesis squared plus left parenthesis negative 2 right parenthesis squared end root end fraction end cell row cell rightwards double arrow text cos end text straight theta equals fraction numerator 6 minus 6 plus 4 over denominator square root of 9 times square root of 9 plus 36 plus 4 end root end fraction end cell row cell rightwards double arrow text cos end text straight theta equals fraction numerator 4 over denominator 3 cross times 7 end fraction end cell row cell rightwards double arrow straight theta equals cos to the power of negative 1 end exponent open parentheses 4 over 21 close parentheses end cell end table end style

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.

Solution 20

begin mathsize 12px style table attributes columnalign left end attributes row cell The text   equation   of   the   plane   parallel   to   ZOX   is   y  =  constant. end text end cell row cell text Given   that   the   y-intercept   is   3. end text end cell row cell text Thus   the   equation   of   the   plane   is   y  =  3. end text end cell end table end style

Question 21

Find the equation of the plane that contains the point (1, -1, 2) and is perpendicular to each of the planes 2x + 3y - 2z = 5 and x + 2y - 3z = 8.

Solution 21

begin mathsize 12px style table attributes columnalign left end attributes row cell text The   equation   of   any   plane   passing   through   end text left parenthesis text 1 , end text minus 1 comma 2 right parenthesis end cell row cell is text   a end text left parenthesis text x end text minus 1 right parenthesis plus straight b left parenthesis straight y plus 1 right parenthesis plus straight c left parenthesis straight z minus 2 right parenthesis equals 0.... left parenthesis 1 right parenthesis end cell row cell Given text   that ,  p end text lane text  ( 1 )  is   perpendicular   to   the   planes end text end cell row cell text 2 x + 3 y end text minus text 2 z  =  5 end text end cell row cell text and end text end cell row cell text x + 2 y end text minus text 3 z  =  8   end text end cell row cell Therefore comma text   we   have , end text end cell row cell text 2 a + 3 b end text minus text 2 c  =  0 end text... text ( 2 ) end text end cell row cell text and   end text end cell row cell text a + 2 b end text minus text 3 c  =  0 end text... text ( 3 ) end text end cell row cell text Solving   equations  ( 2 )  and  ( 3 )  by   cross   multiplication ,  we   have , end text end cell row cell fraction numerator text a end text over denominator text 3 end text cross times left parenthesis negative text 3 end text right parenthesis minus 2 cross times left parenthesis negative 2 right parenthesis end fraction equals fraction numerator text b end text over denominator 1 cross times left parenthesis negative 2 right parenthesis minus 2 cross times left parenthesis negative 3 right parenthesis end fraction equals fraction numerator text c end text over denominator 2 cross times 2 minus 1 cross times 3 end fraction equals straight lambda left parenthesis say right parenthesis end cell row cell rightwards double arrow fraction numerator text a end text over denominator negative 9 plus 4 end fraction equals fraction numerator text b end text over denominator negative 2 plus 6 end fraction equals fraction numerator text c end text over denominator 4 minus 3 end fraction equals straight lambda end cell row cell rightwards double arrow fraction numerator text a end text over denominator negative 5 end fraction equals fraction numerator text b end text over denominator 4 end fraction equals fraction numerator text c end text over denominator 1 end fraction equals straight lambda end cell row cell Thus comma text   we   have , end text end cell row cell text a  = end text minus 5 straight lambda comma straight b equals 4 straight lambda text   and   c  =  end text straight lambda end cell row cell Substituting text   the   above   values   in   equation  ( 1 ),  we   have , end text end cell row cell negative 5 straight lambda left parenthesis text x end text minus 1 right parenthesis plus 4 straight lambda left parenthesis straight y plus 1 right parenthesis plus straight lambda left parenthesis straight z minus 2 right parenthesis equals 0 end cell row cell Since text   end text straight lambda not equal to text 0 ,  we   have , end text end cell row cell negative 5 left parenthesis text x end text minus 1 right parenthesis plus 4 left parenthesis straight y plus 1 right parenthesis plus left parenthesis straight z minus 2 right parenthesis equals 0 end cell row cell rightwards double arrow negative 5 straight x plus 5 plus 4 straight y plus 4 plus straight z minus 2 equals 0 end cell row cell rightwards double arrow negative 5 straight x plus 4 straight y plus straight z plus 7 equals 0 end cell row cell rightwards double arrow 5 straight x minus 4 straight y minus straight z minus 7 equals 0 end cell row cell rightwards double arrow 5 straight x minus 4 straight y minus straight z equals 7 end cell row cell text Thus   the   required   equation   of   the   plane   is   end text 5 straight x minus 4 straight y minus straight z equals 7 end cell end table end style

Question 22

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   the   equation   of   the   plane   passing   through   end text end cell row cell left parenthesis straight a comma straight b comma straight c right parenthesis text   and   parallel   to   the   plane   end text stack text r end text with rightwards arrow on top. left parenthesis straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis equals 2. end cell end table end style

Solution 22

begin mathsize 12px style table attributes columnalign left end attributes row cell Given text   that   the   equation   of   the   required   end text end cell row cell text p end text lane text   is   parallel   to   the   plane end text end cell row cell straight r with rightwards arrow on top times left parenthesis straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis equals 2... left parenthesis 1 right parenthesis end cell row cell therefore Vector text   equation   of   any   plane   parallel   to  ( 1 )  is end text end cell row cell stack text r end text with rightwards arrow on top times left parenthesis straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis text = end text straight k... left parenthesis 2 right parenthesis end cell row cell Since text   the   given   plane   passes   through   end text left parenthesis text a ,  b ,  c end text right parenthesis comma then end cell row cell left parenthesis straight a straight i with hat on top plus straight b straight j with hat on top plus straight c straight k with hat on top right parenthesis times left parenthesis straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis text = end text straight k end cell row cell rightwards double arrow straight a plus straight b plus straight c equals straight k... left parenthesis 3 right parenthesis end cell row blank row cell Substituting text   the   above   value   of   k   in   equation  ( 2 ),  we   have , end text end cell row cell stack text r end text with rightwards arrow on top times left parenthesis straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis text = end text straight a plus straight b plus straight c end cell row cell text Thus   the   required   equation   of   the   plane   is   end text straight x plus straight y plus straight z text   =  end text straight a plus straight b plus straight c end cell end table end style

Question 23

Find the equation of the plane passing through the point (-1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

Solution 23

begin mathsize 12px style table attributes columnalign left end attributes row cell text The   equation   of   any   plane   passing   through   end text left parenthesis negative 1 comma 3 comma 2 right parenthesis end cell row cell is left parenthesis text x  +  end text 1 right parenthesis plus straight b left parenthesis straight y minus 3 right parenthesis plus straight c left parenthesis straight z minus 2 right parenthesis equals 0.... left parenthesis 1 right parenthesis end cell row cell Given text   that ,  end text Plane text  ( 1 )  is   perpendicular   to   the   planes end text end cell row cell text x  +  2 y  +  3 z  =  5 end text end cell row cell text and end text end cell row cell text 3 x  +  3 y end text plus text z  =  0 end text end cell row cell Therefore comma text   we   have , end text end cell row cell text a  +  2 b end text plus 3 text c  =  0 end text... text ( 2 ) end text end cell row cell text and   end text end cell row cell text 3 a  +  3 b end text plus text c  =  0 end text... text ( 3 ) end text end cell row cell text Solving   equations  ( 2 )  and  ( 3 )  by   cross   multiplication ,  we   have , end text end cell row cell fraction numerator text a end text over denominator text 2 end text cross times 1 minus 3 cross times 3 end fraction equals fraction numerator text b end text over denominator 3 cross times 3 minus 1 cross times 1 end fraction equals fraction numerator text c end text over denominator 1 cross times 3 minus 3 cross times 2 end fraction equals straight lambda left parenthesis say right parenthesis end cell row cell rightwards double arrow fraction numerator text a end text over denominator 2 minus 9 end fraction equals fraction numerator text b end text over denominator 9 minus 1 end fraction equals fraction numerator text c end text over denominator 3 minus 6 end fraction equals straight lambda end cell row cell rightwards double arrow fraction numerator text a end text over denominator negative 7 end fraction equals fraction numerator text b end text over denominator 8 end fraction equals fraction numerator text c end text over denominator negative 3 end fraction equals straight lambda end cell row cell Thus comma text   we   have , end text end cell row cell text a  = end text minus 7 straight lambda comma straight b equals 8 straight lambda text   and   c  = end text minus text 3 end text straight lambda end cell row cell Substituting text   the   above   values   in   equation  ( 1 ),  we   have , end text end cell row cell negative 7 straight lambda left parenthesis text x  +  end text 1 right parenthesis plus 8 straight lambda left parenthesis straight y minus 3 right parenthesis minus text 3 end text straight lambda left parenthesis straight z minus 2 right parenthesis equals 0 end cell row cell Since text   end text straight lambda not equal to text 0 ,  we   have , end text end cell row cell negative 7 left parenthesis text x + end text 1 right parenthesis plus 8 left parenthesis straight y minus 3 right parenthesis minus text 3 end text left parenthesis straight z minus 2 right parenthesis equals 0 end cell row cell rightwards double arrow negative 7 straight x minus 7 plus 8 straight y minus 24 minus 3 straight z plus 6 equals 0 end cell row cell rightwards double arrow negative 7 straight x plus 8 straight y minus 3 straight z minus 25 equals 0 end cell row cell rightwards double arrow 7 straight x minus 8 straight y plus 3 straight z plus 25 equals 0 end cell row blank row cell text Thus   the   required   equation   of   the   plane   is   end text 7 straight x minus 8 straight y plus 3 straight z plus 25 equals 0 end cell end table end style

Question 24

Find the vector equation of the plane through the points (2, 1, -1) and (-1, 3, 4) and perpendicular to the plane x - 2y + 4z = 10.

Solution 24

begin mathsize 12px style table attributes columnalign left end attributes row cell text The   equation   of   any   plane   passing   through   end text left parenthesis 2 comma 1 comma negative 1 right parenthesis end cell row cell is text   a end text left parenthesis text x end text minus 2 right parenthesis plus straight b left parenthesis straight y minus 1 right parenthesis plus straight c left parenthesis straight z plus 1 right parenthesis equals 0.... left parenthesis 1 right parenthesis end cell row cell Also comma text   the   above   plane   passes   through   the   point   end text left parenthesis negative 1 comma 3 comma 4 right parenthesis. end cell row cell Thus comma text   equation  ( 1 ),  becomes , end text end cell row cell text a end text left parenthesis negative text 1 end text minus 2 right parenthesis plus straight b left parenthesis 3 minus 1 right parenthesis plus straight c left parenthesis 4 plus 1 right parenthesis equals 0 end cell row cell rightwards double arrow negative 3 text a end text plus 2 straight b plus 5 straight c equals 0... left parenthesis 2 right parenthesis end cell row cell Given text   that ,  end text Plane text  ( 1 )  is   perpendicular   to   the   plane end text end cell row cell text x end text minus text 2 y + 4 z = 10 end text end cell row cell Therefore comma text   we   have , end text end cell row cell text a end text minus text 2 b end text plus 4 text c = 0 end text... text ( 3 ) end text end cell row blank row cell text Solving   equations  ( 2 )  and  ( 3 )  by   cross   multiplication ,  we   have , end text end cell row cell fraction numerator text a end text over denominator text 2 end text cross times 4 minus 5 cross times left parenthesis negative 2 right parenthesis end fraction equals fraction numerator text b end text over denominator 1 cross times 5 minus left parenthesis negative 3 right parenthesis cross times 4 end fraction equals fraction numerator text c end text over denominator left parenthesis negative 3 right parenthesis cross times left parenthesis negative 2 right parenthesis minus 1 cross times 2 end fraction equals straight lambda left parenthesis say right parenthesis end cell row cell rightwards double arrow fraction numerator text a end text over denominator 8 plus 10 end fraction equals fraction numerator text b end text over denominator 5 plus 12 end fraction equals fraction numerator text c end text over denominator 6 minus 2 end fraction equals straight lambda end cell row cell rightwards double arrow fraction numerator text a end text over denominator 18 end fraction equals fraction numerator text b end text over denominator 17 end fraction equals fraction numerator text c end text over denominator 4 end fraction equals straight lambda end cell row cell Thus comma text   we   have , end text end cell row cell text a = end text 18 straight lambda comma straight b equals 17 straight lambda text   and   c = end text 4 straight lambda end cell row cell Substituting text   the   above   values   in   equation  ( 1 ),  we   have , end text end cell row cell 18 straight lambda left parenthesis text x end text minus 2 right parenthesis plus 17 straight lambda left parenthesis straight y minus 1 right parenthesis plus 4 straight lambda left parenthesis straight z plus 1 right parenthesis equals 0 end cell row cell Since text   end text straight lambda not equal to text 0 ,  we   have , end text end cell row cell 18 left parenthesis text x end text minus 2 right parenthesis plus 17 left parenthesis straight y minus 1 right parenthesis plus 4 left parenthesis straight z plus 1 right parenthesis equals 0 end cell row cell rightwards double arrow 18 straight x minus 36 plus 17 straight y minus 17 plus 4 straight z plus 4 equals 0 end cell row cell rightwards double arrow 18 straight x plus 17 straight y plus 4 straight z minus 49 equals 0 end cell row blank row cell text Thus   the   required   equation   of   the   plane   is   end text 18 straight x plus 17 straight y plus 4 straight z minus 49 equals 0 end cell end table end style

Chapter 29 - The plane Excercise Ex. 29.7

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Chapter 29 - The plane Excercise Ex. 29.8

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   the   vector   equation   of   the   plane   passing   end text end cell row cell text through   the   intersection   of   the   planes   end text end cell row cell stack text r end text with rightwards arrow on top. left parenthesis straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis equals 6 text   and   end text stack text r end text with rightwards arrow on top. left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top plus 4 straight k with hat on top right parenthesis equals negative 5 text   and   the   point   end text left parenthesis text 1 ,  1 ,  1 end text right parenthesis. end cell end table end style

Solution 13

begin mathsize 12px style table attributes columnalign left end attributes row cell text The   equation   of   a   plane   passing   through   the   intersection   of end text end cell row cell stack text r end text with rightwards arrow on top times left parenthesis straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis equals 6 text   and   end text stack text r end text with rightwards arrow on top times left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top plus 4 straight k with hat on top right parenthesis equals negative 5 text   is end text end cell row cell left square bracket stack text r end text with rightwards arrow on top times left parenthesis straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis minus 6 right square bracket plus straight lambda left square bracket stack text r end text with rightwards arrow on top times left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top plus 4 straight k with hat on top right parenthesis plus 5 right square bracket equals 0 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times left square bracket left parenthesis 1 plus 2 straight lambda right parenthesis straight i with hat on top plus left parenthesis 1 plus 3 straight lambda right parenthesis straight j with hat on top plus left parenthesis 1 plus 4 straight lambda right parenthesis straight k with hat on top right square bracket equals left parenthesis 6 minus 5 straight lambda right parenthesis... left parenthesis 1 right parenthesis end cell row cell rightwards double arrow left square bracket straight x straight i with hat on top plus straight y straight j with hat on top plus straight z straight k with hat on top right square bracket times left square bracket left parenthesis 1 plus 2 straight lambda right parenthesis straight i with hat on top plus left parenthesis 1 plus 3 straight lambda right parenthesis straight j with hat on top plus left parenthesis 1 plus 4 straight lambda right parenthesis straight k with hat on top right square bracket equals left parenthesis 6 minus 5 straight lambda right parenthesis end cell row cell rightwards double arrow left square bracket straight x left parenthesis 1 plus 2 straight lambda right parenthesis plus straight y left parenthesis 1 plus 3 straight lambda right parenthesis plus straight z left parenthesis 1 plus 4 straight lambda right parenthesis right square bracket equals left parenthesis 6 minus 5 straight lambda right parenthesis... left parenthesis 2 right parenthesis end cell row cell text The   requried   plane   also   passes   through   the   point end text left parenthesis text 1 ,  1 ,  1 end text right parenthesis. end cell row cell text Substiuting   x  =  1 , y  =  1 , z  =  1   in   equation  ( 2 ),  we   have , end text end cell row cell 1 cross times left parenthesis 1 plus 2 straight lambda right parenthesis plus 1 cross times left parenthesis 1 plus 3 straight lambda right parenthesis plus 1 cross times left parenthesis 1 plus 4 straight lambda right parenthesis equals left parenthesis 6 minus 5 straight lambda right parenthesis end cell row cell rightwards double arrow 1 plus 2 straight lambda plus 1 plus 3 straight lambda plus 1 plus 4 straight lambda equals 6 minus 5 straight lambda end cell row cell rightwards double arrow 3 plus 9 straight lambda equals 6 minus 5 straight lambda end cell row cell rightwards double arrow 14 straight lambda equals 6 minus 3 end cell row cell rightwards double arrow 14 straight lambda equals 3 end cell row cell rightwards double arrow straight lambda equals 3 over 14 end cell end table end style

 

begin mathsize 12px style table attributes columnalign left end attributes row cell text Substituting   the   value   end text straight lambda equals 3 over 14 text   in   equation  ( 1 ),  we   have , end text end cell row cell stack text r end text with rightwards arrow on top times open square brackets open parentheses 1 plus 2 open parentheses 3 over 14 close parentheses close parentheses space straight i with hat on top plus open parentheses 1 plus 3 open parentheses 3 over 14 close parentheses close parentheses space straight j with hat on top plus open parentheses 1 plus 4 open parentheses 3 over 14 close parentheses close parentheses space straight k with hat on top close square brackets equals open parentheses 6 minus 5 open parentheses 3 over 14 close parentheses close parentheses end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times open square brackets 20 over 14 straight i with hat on top plus 23 over 14 straight j with hat on top plus 26 over 14 straight k with hat on top close square brackets equals 69 over 14 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times open square brackets 20 straight i with hat on top plus 23 straight j with hat on top plus 26 straight k with hat on top close square brackets equals 69 end cell end table end style

Question 14

Solution 14

Question 15

Find the equation of the plane through the intersection of the planes 3x - y 2z = 4 and x + y + z = 2 and the point (2, 2, 1).

Solution 15

  

Question 16

Find the vector equation of the plane through the line of intersection of the plane x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x - y + z = 0.

Solution 16

  

Question 17

Find the equation of the plane passing through (a, b, c) and parallel to the plane

  

Solution 17

  

Chapter 29 - The plane Excercise Ex. 29.9

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9
Question 10

Solution 10

Question 11

Find the distance between the point (7, 2, 4) and the plane determined by the points A (2, 5, -3), B (-2, -3, 5) and C (5, 3, -3)

Solution 11

begin mathsize 12px style table attributes columnalign left end attributes row cell text The   equation   of   any   plane   passing   through   A end text left parenthesis 2 comma 5 comma negative 3 right parenthesis end cell row cell is text   a end text left parenthesis text x end text minus 2 right parenthesis plus straight b left parenthesis straight y minus 5 right parenthesis plus straight c left parenthesis straight z plus 3 right parenthesis equals 0.... left parenthesis 1 right parenthesis end cell row cell The text   above   plane   passes   through   the   point   B end text left parenthesis negative text 2 , end text minus text 3 , 5 end text right parenthesis end cell row cell and text   hence ,  we   have , end text end cell row cell straight a left parenthesis negative text 2 end text minus 2 right parenthesis plus straight b left parenthesis negative 3 minus 5 right parenthesis plus straight c left parenthesis 5 plus 3 right parenthesis equals 0 end cell row cell rightwards double arrow negative 4 straight a minus 8 straight b plus 8 straight c equals 0... left parenthesis 2 right parenthesis end cell row cell Again text   the   required   plane   passes   through   the   point   C end text left parenthesis text 5 , 3 , end text minus 3 right parenthesis end cell row cell and text   hence ,  we   have , end text end cell row cell straight a left parenthesis 5 minus 2 right parenthesis plus straight b left parenthesis 3 minus 5 right parenthesis plus straight c left parenthesis negative 3 plus 3 right parenthesis equals 0 end cell row cell rightwards double arrow 3 straight a minus 2 straight b plus 0 straight c equals 0... left parenthesis 3 right parenthesis end cell row cell Solving text   equations  ( 2 )  and  ( 3 )  by   cross   multiplication ,  we   have , end text end cell row cell fraction numerator text a end text over denominator left parenthesis negative text 8 end text right parenthesis cross times 0 minus left parenthesis negative 2 right parenthesis cross times 8 end fraction equals fraction numerator straight b over denominator 3 cross times 8 minus left parenthesis negative 4 right parenthesis cross times 0 end fraction equals fraction numerator straight c over denominator left parenthesis negative 4 right parenthesis cross times left parenthesis negative 2 right parenthesis minus 3 cross times left parenthesis negative 8 right parenthesis end fraction equals straight lambda left parenthesis say right parenthesis end cell row cell rightwards double arrow fraction numerator text a end text over denominator 0 plus 16 end fraction equals fraction numerator straight b over denominator 24 plus 0 end fraction equals fraction numerator straight c over denominator 8 plus 24 end fraction equals straight lambda end cell row cell rightwards double arrow fraction numerator text a end text over denominator 16 end fraction equals straight b over 24 equals straight c over 32 equals straight lambda end cell row cell rightwards double arrow fraction numerator text a end text over denominator 2 end fraction equals straight b over 3 equals straight c over 4 equals straight lambda end cell row cell rightwards double arrow straight a equals 2 straight lambda comma straight b equals 3 straight lambda text   and   c = 4 end text straight lambda end cell row cell Substituting text   the   above   values   in   equation  ( 1 ),  we   have , end text end cell row cell 2 straight lambda left parenthesis text x end text minus 2 right parenthesis plus 3 straight lambda left parenthesis straight y minus 5 right parenthesis plus text 4 end text straight lambda left parenthesis straight z plus 3 right parenthesis equals 0 end cell row cell Since text   end text straight lambda not equal to text 0 ,  we   have , end text end cell row cell 2 left parenthesis text x end text minus 2 right parenthesis plus 3 left parenthesis straight y minus 5 right parenthesis plus text 4 end text left parenthesis straight z plus 3 right parenthesis equals 0 end cell row cell rightwards double arrow 2 straight x minus 4 plus 3 straight y minus 15 plus text 4 z + 12 end text equals 0 end cell row cell rightwards double arrow 2 straight x plus 3 straight y plus text 4 z end text minus text 7 end text equals 0 end cell end table end style

 

begin mathsize 12px style table attributes columnalign left end attributes row cell Thus text   the   equation   of   the   plane   is   end text end cell row cell 2 straight x plus 3 straight y plus text 4 z end text minus text 7 end text equals 0 end cell row cell The text   distance   from   the   point   P end text left parenthesis text 7 , 2 , 4 end text right parenthesis text   to   the   plane   is end text end cell row cell text d = end text open vertical bar fraction numerator text ax end text subscript text 1 end text end subscript plus b y subscript 1 plus c z subscript 1 plus straight d over denominator square root of straight a squared plus straight b squared plus straight c squared end root end fraction close vertical bar end cell row cell therefore text Distance ,  d = end text open vertical bar fraction numerator 2 straight x plus 3 straight y plus text 4 z end text minus text 7 end text over denominator square root of 2 squared plus 3 squared plus 4 squared end root end fraction close vertical bar end cell row cell rightwards double arrow straight d subscript left parenthesis 7 comma 2 comma 4 right parenthesis end subscript equals open vertical bar fraction numerator 2 cross times 7 plus 3 cross times 2 plus text 4 end text cross times text 4 end text minus text 7 end text over denominator square root of 2 squared plus 3 squared plus 4 squared end root end fraction close vertical bar end cell row cell rightwards double arrow straight d subscript left parenthesis 7 comma 2 comma 4 right parenthesis end subscript equals open vertical bar fraction numerator 29 over denominator square root of 29 end fraction close vertical bar end cell row cell rightwards double arrow straight d subscript left parenthesis 7 comma 2 comma 4 right parenthesis end subscript equals square root of 29 text   units end text end cell end table end style

Question 12

A plane makes intercepts -6, 3, 4 respectively on the coordinate axes. Find the length of the perpendicular from the origin on it.

Solution 12

begin mathsize 12px style table attributes columnalign left end attributes row cell Given text   that   a   plane   is   making   intercepts   end text minus 6 comma 3 text   and   4 end text end cell row cell text respectively   on   the   coordinate   axes. end text end cell row cell text Thus   the   equation   of   the   plane   is end text end cell row cell fraction numerator text x end text over denominator negative text 6 end text end fraction plus straight y over 3 plus straight z over 4 equals 1... left parenthesis 1 right parenthesis end cell row cell We text   need   to   find   the   length   of   the   perpendicular end text end cell row cell text from   the   origin   on   the   plane. end text end cell row cell text If   the   plane   end text fraction numerator text x end text over denominator straight a end fraction plus straight y over straight b plus straight z over straight c equals 1 text   is   at   a   distance  ' p ',  then end text end cell row cell fraction numerator text 1 end text over denominator text p end text to the power of text 2 end text end exponent end fraction equals 1 over straight a squared plus 1 over straight b squared plus 1 over straight c squared... left parenthesis 2 right parenthesis end cell row cell Comparing text   equation  ( 1 )  with   the   end text end cell row cell text general   equation ,  we   get , end text end cell row cell text a = end text minus 6 comma straight b equals 3 text   and   c = 4 end text end cell row cell text Thus ,  equation  ( 2 )  becomes , end text end cell row cell fraction numerator text 1 end text over denominator text p end text to the power of text 2 end text end exponent end fraction equals 1 over left parenthesis negative 6 right parenthesis squared plus 1 over 3 squared plus 1 over 4 squared end cell row cell rightwards double arrow fraction numerator text 1 end text over denominator text p end text to the power of text 2 end text end exponent end fraction equals fraction numerator 4 plus 16 plus 9 over denominator 144 end fraction end cell row cell rightwards double arrow fraction numerator text 1 end text over denominator text p end text to the power of text 2 end text end exponent end fraction equals 29 over 144 end cell row cell rightwards double arrow text p end text to the power of text 2 end text end exponent equals 144 over 29 end cell row cell rightwards double arrow text p end text equals fraction numerator 12 over denominator square root of 29 end fraction text   units end text end cell end table end style

 

Chapter 29 - The plane Excercise Ex. 29.10

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Chapter 29 - The plane Excercise Ex. 29.11

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

begin mathsize 12px style table attributes columnalign left end attributes row cell Find space the space vector space and space cartesian space forms space of space the space equation space end cell row cell of space the space plane space passing space through space the space point space left parenthesis 1 comma space 2 comma space minus 4 right parenthesis space and space end cell row cell parallel space to space the space lines space straight r with rightwards arrow on top equals left parenthesis straight i with hat on top plus 2 straight j with hat on top minus 4 straight k with hat on top right parenthesis plus straight lambda left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top plus 6 straight k with hat on top right parenthesis space and end cell row cell straight r with rightwards arrow on top equals left parenthesis straight i with hat on top minus 3 straight j with hat on top plus 5 straight k with hat on top right parenthesis plus straight mu left parenthesis straight i with hat on top plus straight j with hat on top minus straight k with hat on top right parenthesis .  Also comma space find space the space distance space of space end cell row cell the space point space left parenthesis 9 ,- 8 ,- 10 right parenthesis space from space the space plane space thus space obtained. end cell end table end style

Solution 18

begin mathsize 12px style table attributes columnalign left end attributes row cell text The   plane   passes   through   the   point   end text stack text a end text with rightwards arrow on top open parentheses 1 comma space 2 comma space minus 4 close parentheses end cell row cell text A   vector   in   a   direction   perpendicular   to   end text end cell row cell stack text r end text with rightwards arrow on top equals open parentheses straight i with hat on top plus 2 straight j with hat on top minus 4 straight k with hat on top close parentheses plus straight lambda open parentheses 2 straight i with hat on top plus 3 straight j with hat on top plus 6 straight k with hat on top close parentheses text   and   end text stack text r end text with rightwards arrow on top equals open parentheses straight i with hat on top minus 3 straight j with hat on top plus 5 straight k with hat on top close parentheses plus straight mu open parentheses straight i with hat on top plus straight j with hat on top minus straight k with hat on top close parentheses end cell row cell text is   end text stack text n end text with rightwards arrow on top equals left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top plus 6 straight k with hat on top right parenthesis cross times left parenthesis straight i with hat on top plus straight j with hat on top minus straight k with hat on top right parenthesis end cell row cell rightwards double arrow stack text n end text with rightwards arrow on top equals open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 2 3 6 row 1 1 cell negative 1 end cell end table close vertical bar equals negative 9 straight i with hat on top plus 8 straight j with hat on top minus straight k with hat on top end cell row cell text Equation   of   the   plane   is   end text open parentheses straight r with rightwards arrow on top minus straight a with rightwards arrow on top close parentheses. straight n with rightwards arrow on top equals 0 end cell row cell open parentheses straight r with rightwards arrow on top minus open parentheses straight i with hat on top plus 2 straight j with hat on top minus 4 straight k with hat on top close parentheses close parentheses. open parentheses negative 9 straight i with hat on top plus 8 straight j with hat on top minus straight k with hat on top close parentheses equals 0 end cell row cell rightwards double arrow straight r with rightwards arrow on top. open parentheses negative 9 straight i with hat on top plus 8 straight j with hat on top minus straight k with hat on top close parentheses equals 11 end cell row cell text Substituting   end text straight r with rightwards arrow on top equals straight x straight i with hat on top plus straight y straight j with hat on top plus straight z straight k with hat on top comma text   we   get   the   Cartesian   form   as end text end cell row cell negative 9 straight x plus 8 straight y minus straight z equals 11 end cell row cell text The   distance   of   the   point   end text left parenthesis text 9 , end text minus text 8 , end text minus text 10 end text right parenthesis text   from   the   plane end text end cell row cell text = end text open vertical bar fraction numerator negative 9 left parenthesis 9 right parenthesis plus 8 left parenthesis negative 8 right parenthesis minus left parenthesis negative 10 right parenthesis minus 11 over denominator square root of 9 squared plus 8 squared plus 1 squared end root end fraction close vertical bar equals fraction numerator 146 over denominator square root of 146 end fraction equals square root of 146 end cell end table end style

Question 19

Solution 19

Question 20

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   the   coordinates   of   the   point   where   the   line   end text end cell row cell fraction numerator text x -2 end text over denominator text 3 end text end fraction equals fraction numerator straight y plus 1 over denominator 4 end fraction equals fraction numerator straight z minus 2 over denominator 2 end fraction text   intersects   the   plane   x-y + z -5 = 0 .  end text end cell row cell text Also ,  find   the   angle   between   the   line   and   the   plane. end text end cell end table end style

Solution 20

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   the   coordinates   of   the   point   where   the   line   end text end cell row cell fraction numerator text x -2 end text over denominator text 3 end text end fraction equals fraction numerator straight y plus 1 over denominator 4 end fraction equals fraction numerator straight z minus 2 over denominator 2 end fraction text = r end text end cell row cell rightwards double arrow straight x equals 3 straight r plus 2 comma straight y equals 4 straight r minus 1 comma straight z equals 2 straight r plus 2 end cell row cell text Substituting   in   the   equation   of   the   plane   x  -  y  +  z  -  5  =  0 ,  end text end cell row cell text we   get   end text left parenthesis 3 straight r plus 2 right parenthesis minus left parenthesis 4 straight r minus 1 right parenthesis plus left parenthesis 2 straight r plus 2 right parenthesis minus 5 equals 0 end cell row cell rightwards double arrow straight r equals 0 end cell row cell rightwards double arrow straight x equals 2 comma straight y equals negative 1 comma straight z equals 2 end cell row cell text Direction   ratios   of   the   line   are   end text 3 comma 4 comma 2 end cell row cell text Direction   ratios   of   a   line   perpendicular   to   the   plane   are   end text 1 comma negative 1 comma 1 end cell row cell sinθ equals fraction numerator 3 cross times 1 plus 4 cross times negative 1 plus 2 cross times 1 over denominator square root of 9 plus 16 plus 4 end root square root of 1 plus 1 plus 1 end root end fraction equals fraction numerator 1 over denominator square root of 87 end fraction end cell row cell straight theta equals sin to the power of negative 1 end exponent fraction numerator 1 over denominator square root of 87 end fraction end cell end table end style

Question 21

Solution 21

Question 22

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   the   angle   between   the   line   end text fraction numerator text x + 1 end text over denominator text 2 end text end fraction equals straight y over 3 equals fraction numerator straight z minus 3 over denominator 6 end fraction text   end text end cell row cell text and   the   plane   end text 10 straight x plus 2 straight y minus 11 straight z text   end text equals text   end text 3 end cell end table end style

Solution 22

begin mathsize 12px style table attributes columnalign left end attributes row cell text Direction   ratios   of   the   line   end text fraction numerator text x + 1 end text over denominator text 2 end text end fraction equals straight y over 3 equals fraction numerator straight z minus 3 over denominator 6 end fraction text   end text end cell row cell text are   end text left parenthesis 2 comma 3 comma 6 right parenthesis end cell row cell text Direction   ratio   of   a   line   perpendicular   to   the   plane   end text end cell row cell 10 straight x plus 2 straight y minus 11 straight z text   end text equals text   end text 3 text   are   end text 10 comma 2 comma negative 11 end cell row cell text If   end text straight theta text   is   the   angle   between   the   line   and   the   plane ,  then   end text end cell row cell sinθ equals fraction numerator 2 cross times 10 plus 3 cross times 2 plus 6 cross times negative 11 over denominator square root of 2 squared plus 3 squared plus 6 squared end root square root of 10 squared plus 2 squared plus 11 squared end root end fraction equals negative fraction numerator 40 over denominator square root of 49 square root of 225 end fraction equals negative fraction numerator 40 over denominator 7 cross times 15 end fraction equals negative 8 over 21 end cell row cell rightwards double arrow straight theta equals sin to the power of negative 1 end exponent open parentheses negative 8 over 21 close parentheses end cell end table end style

Question 23

Solution 23

Question 24

Solution 24

Question 25

Find the equation of the plane passing through the points

(-1, 2, 0),(2, 2, -1) and parallel to the line

  

Solution 25

  

Chapter 29 - The plane Excercise Ex. 29.12

Question 1

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the yz-plane.

Solution 1

begin mathsize 12px style table attributes columnalign left end attributes row cell text Direction   ratios   of   the   given   line   are end text end cell row cell left parenthesis 5 minus 3 comma 1 minus 4 comma 6 minus 1 right parenthesis equals left parenthesis 2 comma negative 3 comma 5 right parenthesis end cell row cell text Hence ,  equation   of   the   line   is   end text end cell row cell fraction numerator straight x minus 5 over denominator 2 end fraction equals fraction numerator straight y minus 1 over denominator negative 3 end fraction equals fraction numerator straight z minus 6 over denominator 5 end fraction equals straight r end cell row cell rightwards double arrow straight x equals 2 straight r plus 5 comma straight y equals negative 3 straight r plus 1 comma straight z equals 5 straight r plus 6 end cell row cell text For   any   point   on   the   end text yz minus text plane   end text straight x equals 0 end cell row cell rightwards double arrow 2 straight r plus 5 equals 0 rightwards double arrow straight r equals negative 5 over 2 end cell row cell straight y equals negative 3 left parenthesis negative 5 over 2 right parenthesis plus 1 equals 17 over 2 end cell row cell straight z equals 5 left parenthesis negative 5 over 2 right parenthesis plus 6 equals negative 13 over 2 end cell row cell text Hence ,  the   coordinates   of   the   point   are   end text open parentheses 0 comma 17 over 2 comma negative 13 over 2 close parentheses. end cell end table end style

Question 2

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the zx-plane.

Solution 2

begin mathsize 12px style table attributes columnalign left end attributes row cell text Direction   ratios   of   the   given   line   are end text end cell row cell left parenthesis 5 minus 3 comma 1 minus 4 comma 6 minus 1 right parenthesis equals left parenthesis 2 comma negative 3 comma 5 right parenthesis end cell row cell text Hence ,  equation   of   the   line   is   end text end cell row cell fraction numerator straight x minus 5 over denominator 2 end fraction equals fraction numerator straight y minus 1 over denominator negative 3 end fraction equals fraction numerator straight z minus 6 over denominator 5 end fraction equals straight r end cell row cell rightwards double arrow straight x equals 2 straight r plus 5 comma straight y equals negative 3 straight r plus 1 comma straight z equals 5 straight r plus 6 end cell row cell text For   any   point   on   end text zx minus text plane   end text straight y equals 0 end cell row cell rightwards double arrow negative 3 straight r plus 1 equals 0 rightwards double arrow straight r equals 1 third end cell row cell straight x equals 2 open parentheses 1 third close parentheses plus 5 equals 17 over 3 end cell row cell straight z equals 5 open parentheses 1 third close parentheses plus 6 equals 23 over 3 end cell row cell text Hence ,  the   coordinates   of   the   point   are   end text open parentheses 17 over 3 comma 0 comma 23 over 3 close parentheses. end cell end table end style

Question 3

F i n d space t h e space c o o r d i n a t e s space o f space t h e space p o i n t space w h e r e space t h e space l i n e space t h r o u g h space open parentheses 3 comma negative 4 comma negative 5 close parentheses space a n d
open parentheses 2 comma negative 3 comma 1 close parentheses space c r o s s e s space t h e space p l a n e space 2 x plus y plus z equals 7

Solution 3

L e t space t h e space c o o r d i n a t e s space o f space t h e space p o i n t s space A space a n d space B space b e space
open parentheses 3 comma negative 4 comma negative 5 close parentheses space a n d space open parentheses 2 comma negative 3 comma 1 close parentheses space r e p e c t i v e l y.
E q u a t i o n space o f space t h e space l i n e space j o i n i n g space t h e space p o i n t s space open parentheses x subscript 1 comma y subscript 1 comma z subscript 1 close parentheses space a n d space open parentheses x subscript 2 comma y subscript 2 comma z subscript 2 close parentheses space i s
fraction numerator x minus x subscript 1 over denominator x subscript 2 minus x subscript 1 end fraction equals fraction numerator y minus y subscript 1 over denominator y subscript 2 minus y subscript 1 end fraction equals fraction numerator z minus z subscript 1 over denominator z subscript 2 minus z subscript 1 end fraction equals r comma space w h e r e space r space i s space s o m e space c o n s tan t.
T h u s space e q u a t i o n space o f space A B space i s
fraction numerator x minus 3 over denominator 2 minus 3 end fraction equals fraction numerator y minus open parentheses negative 4 close parentheses over denominator open parentheses negative 3 close parentheses minus open parentheses negative 4 close parentheses end fraction equals fraction numerator z minus open parentheses negative 5 close parentheses over denominator 1 minus open parentheses negative 5 close parentheses end fraction equals r
rightwards double arrow fraction numerator x minus 3 over denominator negative 1 end fraction equals fraction numerator y plus 4 over denominator 1 end fraction equals fraction numerator z plus 5 over denominator 6 end fraction equals r
A n y space p o i n t space o n space t h e space l i n e space A B space i s space o f space t h e space f o r m
minus r plus 3 comma space r minus 4 comma space 6 r minus 5
L e t space P space b e space t h e space p o i n t space o f space i n t e r s e c t i o n space o f space t h e space l i n e space A B space a n d space t h e space p l a n e space 2 x plus y plus z equals 7
T h u s comma space w e space h a v e comma
2 open parentheses negative r plus 3 close parentheses plus space r minus 4 plus 6 r minus 5 equals 7
rightwards double arrow negative 2 r plus 6 plus space r minus 4 plus 6 r minus 5 equals 7
rightwards double arrow 5 r equals 10
rightwards double arrow r equals 2
S u b s t i t u t i n g space t h e space v a l u e space o f space r space i n space minus r plus 3 comma space r minus 4 comma space 6 r minus 5 comma space t h e space c o o r d i n a t e s space o f space P space a r e colon
open parentheses negative 2 plus 3 comma space 2 minus 4 comma space 6 cross times 2 minus 5 close parentheses equals open parentheses 1 comma negative 2 comma 7 close parentheses

Question 4

F i n d space t h e space d i s tan c e space o f space t h e space p o i n t space open parentheses negative 1 comma negative 5 comma negative 10 close parentheses space f r o m space t h e space p o i n t space o f space i n t e r s e c t i o n
o f space t h e space l i n e space r with rightwards arrow on top equals open parentheses 2 i with hat on top minus j with hat on top plus 2 k with overparenthesis on top close parentheses plus lambda open parentheses 3 i with hat on top plus 4 j with hat on top plus 2 k with hat on top close parentheses space a n d space t h e space p l a n e
r with rightwards arrow on top times open parentheses i with hat on top minus j with hat on top plus k with hat on top close parentheses equals 5

Solution 4

Question 5

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   the   distance   of   the   point   end text left parenthesis text 2 , 12 , 5 end text right parenthesis text   from   the   point   end text end cell row cell text of   intersection   of   the   line   end text stack text r end text with rightwards arrow on top equals 2 straight i with hat on top minus 4 straight j with hat on top plus 2 straight k with hat on top plus straight lambda open parentheses 3 straight i with hat on top plus 4 straight j with hat on top plus 2 straight k with hat on top close parentheses end cell row cell text and   the   plane   end text stack text r end text with rightwards arrow on top. open parentheses straight i with hat on top minus 2 straight j with hat on top plus straight k with hat on top close parentheses equals 0. end cell end table end style

Solution 5

begin mathsize 12px style table attributes columnalign left end attributes row cell text To   find   the   point   of   intersection   of   the   line   end text end cell row cell stack text r end text with rightwards arrow on top equals 2 straight i with hat on top minus 4 straight j with hat on top plus 2 straight k with hat on top plus straight lambda open parentheses 3 straight i with hat on top plus 4 straight j with hat on top plus 2 straight k with hat on top close parentheses text   and   the   plane   end text stack text r end text with rightwards arrow on top. open parentheses straight i with hat on top minus 2 straight j with hat on top plus straight k with hat on top close parentheses equals 0 comma end cell row cell text we   substitute   end text stack text r end text with rightwards arrow on top text   of   line   in   the   plane. end text end cell row cell open square brackets 2 straight i with hat on top minus 4 straight j with hat on top plus 2 straight k with hat on top plus straight lambda open parentheses 3 straight i with hat on top plus 4 straight j with hat on top plus 2 straight k with hat on top close parentheses close square brackets. open parentheses straight i with hat on top minus 2 straight j with hat on top plus straight k with hat on top close parentheses equals 0 end cell row cell rightwards double arrow open square brackets open parentheses 2 plus 3 straight lambda close parentheses straight i with hat on top plus open parentheses negative 4 plus 4 straight lambda close parentheses straight j with hat on top plus open parentheses 2 plus 2 straight lambda close parentheses straight k with hat on top close square brackets. open parentheses straight i with hat on top minus 2 straight j with hat on top plus straight k with hat on top close parentheses equals 0 end cell row cell rightwards double arrow 2 plus 3 straight lambda plus 8 minus 8 straight lambda plus 2 plus 2 straight lambda equals 0 end cell row cell rightwards double arrow 3 straight lambda equals 12 rightwards double arrow straight lambda equals 4 end cell row cell stack text r end text with rightwards arrow on top equals 2 straight i with hat on top minus 4 straight j with hat on top plus 2 straight k with hat on top plus 4 open parentheses 3 straight i with hat on top plus 4 straight j with hat on top plus 2 straight k with hat on top close parentheses equals 14 straight i with hat on top plus 12 straight j with hat on top plus 10 straight k with hat on top end cell row cell text Hence ,  the   distance   of   the   point   2 end text straight i with hat on top plus 12 straight j with hat on top plus 5 straight k with hat on top text   from   end text 14 straight i with hat on top plus 12 straight j with hat on top plus 10 straight k with hat on top text   is end text end cell row cell square root of left parenthesis 14 minus 2 right parenthesis squared plus left parenthesis 12 minus 12 right parenthesis squared plus left parenthesis 10 minus 5 right parenthesis squared end root equals square root of 12 squared plus 5 squared end root equals square root of 169 equals 13 end cell end table end style

Question 6

Find the distance of the point P (-1, -5, -10) from the point of intersection of the line joining the points A (2, -1, 2) and B (5, 3, 4) with the plane x - y + z = 5.

Solution 6

begin mathsize 12px style table attributes columnalign left end attributes row cell text Equation   of   the   line   through   the   points   A end text left parenthesis 2 comma negative 1 comma 2 right parenthesis end cell row cell text and   B end text left parenthesis 5 comma 3 comma 4 right parenthesis text   is   end text fraction numerator straight x minus 2 over denominator 5 minus 2 end fraction equals fraction numerator straight y plus 1 over denominator 3 plus 1 end fraction equals fraction numerator straight z minus 2 over denominator 4 minus 2 end fraction equals straight r end cell row cell rightwards double arrow fraction numerator straight x minus 2 over denominator 3 end fraction equals fraction numerator straight y plus 1 over denominator 4 end fraction equals fraction numerator straight z minus 2 over denominator 2 end fraction equals straight r end cell row cell rightwards double arrow straight x equals 3 straight r plus 2 comma straight y equals 4 straight r minus 1 comma straight z equals 2 straight r plus 2 end cell row cell text Substituting   these   in   the   plane   equation   we   get end text end cell row cell left parenthesis 3 straight r plus 2 right parenthesis minus left parenthesis 4 straight r minus 1 right parenthesis plus left parenthesis 2 straight r plus 2 right parenthesis equals 5 end cell row cell rightwards double arrow straight r equals 0 end cell row cell rightwards double arrow straight x equals 2 comma straight y equals negative 1 comma straight z equals 2 end cell row cell text Distance   of   end text left parenthesis 2 comma negative 1 comma 2 right parenthesis text   from   end text left parenthesis negative 1 comma negative 5 comma negative 10 right parenthesis text   is end text end cell row cell equals square root of left parenthesis 2 minus left parenthesis negative 1 right parenthesis right parenthesis squared plus left parenthesis negative 1 minus left parenthesis negative 5 right parenthesis right parenthesis squared plus left parenthesis 2 minus left parenthesis negative 10 right parenthesis right parenthesis squared end root equals square root of 3 squared plus 4 squared plus 12 squared end root end cell row cell equals square root of 169 equals 13 end cell end table end style

Question 7

Find the distance of the point P(3, 4,4) from the point, where the line joining the points A(3, -4, -5) and B (2, -3, 1) intersects the plane 2x + y + z =7.

Solution 7

  

Chapter 29 - The plane Excercise Ex. 29.13

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Show space that space the space plane space whose space vector space equation space is space straight r with rightwards arrow on top. left parenthesis straight i with hat on top plus 2 straight j with hat on top minus straight k with hat on top right parenthesis space equals 3 space contains space the space line
whose space vector space equation space is space straight r with rightwards arrow on top equals straight i with hat on top plus straight j with hat on top plus straight lambda left parenthesis 2 straight i with hat on top plus straight j with hat on top plus 4 straight k with hat on top right parenthesis.

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Let space the space equation space of space the space plane space be space straight x over straight a plus straight y over straight b plus straight z over straight c equals 1........ left parenthesis straight i right parenthesis
Plane space is space passing space through space left parenthesis 3 comma 4 comma 2 right parenthesis space and space left parenthesis 7 comma 0 comma 6 right parenthesis
3 over straight a plus 4 over straight b plus 2 over straight c equals 1
7 over straight a plus 0 over straight b plus 6 over straight c equals 1
Required space plane space is space perpendicular space to space 2 straight x minus 5 straight y minus 15 equals 0
2 over straight a plus fraction numerator negative 5 over denominator straight b end fraction plus 0 over straight c equals 0
rightwards double arrow 2 straight b equals 5 straight a
therefore straight b equals 2.5 straight a
3 over straight a plus fraction numerator 4 over denominator 2.5 straight a end fraction plus 2 over straight c equals 1
7 over straight a plus 6 over straight c equals 1
Solving space the space above space 2 space equations comma
straight a equals 3.4 equals 17 over 5 comma space straight b equals space 8.5 equals 17 over 2 space and space straight c equals fraction numerator negative 34 over denominator 6 end fraction equals negative 17 over 3
Substituting space the space values space in space left parenthesis straight i right parenthesis
fraction numerator straight x over denominator 17 over 5 end fraction plus fraction numerator straight y over denominator 17 over 2 end fraction plus fraction numerator straight z over denominator negative 17 over 3 end fraction equals 1
rightwards double arrow fraction numerator 5 straight x over denominator 17 end fraction plus fraction numerator 2 straight y over denominator 17 end fraction minus fraction numerator 3 straight z over denominator 17 end fraction equals 1
rightwards double arrow 5 straight x plus 2 straight y minus 3 straight z equals 17
rightwards double arrow left parenthesis straight x straight i with hat on top plus straight y straight j with hat on top plus straight z straight k with hat on top right parenthesis. left parenthesis 5 straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis equals 17
rightwards double arrow straight r with rightwards arrow on top. left parenthesis 5 straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis equals 17
Vector space equation space of space the space plane space is space straight r with rightwards arrow on top. left parenthesis 5 straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis equals 17.
The space line space passes space through space straight B left parenthesis 1 comma 3 comma negative 2 right parenthesis.
5 left parenthesis 1 right parenthesis plus 2 left parenthesis 3 right parenthesis minus 3 left parenthesis negative 2 right parenthesis equals 17
The space point space straight B space lies space on space the space plane.
therefore The space line space straight r with rightwards arrow on top equals straight i with hat on top plus 3 straight j with hat on top minus 2 straight k with hat on top plus straight lambda left parenthesis straight i with hat on top minus straight j with hat on top plus straight k with hat on top right parenthesis space lies space on space the space plane space straight r with rightwards arrow on top. left parenthesis 5 straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis equals 17.

Question 9

Solution 9

Question 10

F i n d space t h e space c o o r d i n a t e s space o f space t h e space p o i n t space w h e r e space t h e space l i n e
fraction numerator x minus 2 over denominator 3 end fraction equals fraction numerator y plus 1 over denominator 4 end fraction equals fraction numerator z minus 2 over denominator 2 end fraction space i n t e r s e c t s space t h e space p l a n e space x minus y plus z minus 5 equals 0. space A l s o comma space f i n d
t h e space a n g l e space b e t w e e n space t h e space l i n e space a n d space t h e space p l a n e.

Solution 10

A n y space p o i n t space o n space t h e space l i n e
fraction numerator x minus 2 over denominator 3 end fraction equals fraction numerator y plus 1 over denominator 4 end fraction equals fraction numerator z minus 2 over denominator 2 end fraction space equals k
i s space o f space t h e space f o r m comma space open parentheses 3 k plus 2 comma space 4 k minus 1 comma space 2 k plus 2 close parentheses.
I f space t h e space p o i n t space P open parentheses 3 k plus 2 comma space 4 k minus 1 comma space 2 k plus 2 close parentheses space l i e s space i n space t h e space p l a n e space x minus y plus z minus 5 equals 0 comma space w e space h a v e comma
open parentheses 3 k plus 2 close parentheses minus open parentheses 4 k minus 1 close parentheses plus open parentheses 2 k plus 2 close parentheses minus 5 equals 0
rightwards double arrow 3 k plus 2 minus 4 k plus 1 plus 2 k plus 2 minus 5 equals 0
rightwards double arrow k equals 0
T h u s comma space t h e space c o o r d i n a t e s space o f space t h e space p o i n t space o f space i n t e r s e c t i o n space o f space t h e space l i n e space a n d
t h e space p l a n e space a r e colon space P open parentheses 3 cross times 0 plus 2 comma space 4 cross times 0 minus 1 comma space 2 cross times 0 plus 2 close parentheses equals P open parentheses 2 comma negative 1 comma 2 close parentheses

L e t space theta space b e space t h e space a n g l e space b e t w e e n space t h e space l i n e space a n d space t h e space p l a n e.
T h u s comma
sin theta equals fraction numerator a l plus b m plus c n over denominator square root of a squared plus b squared plus c squared end root square root of l squared plus m squared plus n squared end root end fraction comma space w h e r e comma space l comma m space a n d space n space a r e space t h e space d i r e c t i o n
r a t i o s space o f space t h e space l i n e space a n d space a comma b space a n d space c space a r e space t h e space d i r e c t i o n space r a t i o s space o f space t h e space n o r m a l
t o space t h e space p l a n e.
H e r e comma space l equals 3 comma space m equals 4 comma space n equals 2 comma space a equals 1 comma space b equals negative 1 comma space a n d space c equals 1
H e n c e comma
sin theta equals fraction numerator 1 cross times 3 plus open parentheses negative 1 close parentheses cross times 4 plus 1 cross times 2 over denominator square root of 1 squared plus open parentheses negative 1 close parentheses squared plus 1 squared end root square root of 3 squared plus 4 squared plus 2 squared end root end fraction
rightwards double arrow sin theta equals fraction numerator 1 over denominator square root of 3 square root of 29 end fraction
rightwards double arrow theta equals sin to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator square root of 3 square root of 29 end fraction close parentheses

Question 11

F i n d space t h e space v e c t o r space e q u a t i o n space o f space t h e space p l a n e space p a s sin g space t h r o u g h space t h r e e space p o i n t s
w i t h space p o s i t i o n space v e c t o r s space i with hat on top plus j with hat on top minus 2 k with hat on top comma space 2 i with hat on top minus j with hat on top plus k with hat on top space a n d space i with hat on top plus 2 j with hat on top plus k with hat on top. space A l s o comma space f i n d space t h e space
c o o r d i n a t e s space o f space t h e space p o i n t space o f space i n t e r s e c t i o n space o f space t h i s space p l a n e space a n d space
t h e space l i n e space r with rightwards arrow on top equals 3 i with hat on top minus j with hat on top minus k with hat on top plus lambda open parentheses 2 i with hat on top minus 2 j with hat on top plus k with hat on top close parentheses.

Solution 11

L e t space A comma space B space a n d space C space b e space t h r e e space p o i n t s space w i t h space p o s i t i o n space v e c t o r s space
i with hat on top plus j with hat on top minus 2 k with hat on top comma space 2 i with hat on top minus j with hat on top plus k with hat on top space a n d space i with hat on top plus 2 j with hat on top plus k with hat on top. space
T h u s comma space stack A B with rightwards arrow on top equals b with rightwards arrow on top minus a with rightwards arrow on top equals open parentheses 2 i with hat on top minus j with hat on top plus k with hat on top close parentheses minus open parentheses i with hat on top plus j with hat on top minus 2 k with hat on top close parentheses equals i with hat on top minus 2 j with hat on top plus 3 k with hat on top
stack A C with rightwards arrow on top equals c with rightwards arrow on top minus a with rightwards arrow on top equals open parentheses i with hat on top plus 2 j with hat on top plus k with hat on top close parentheses minus open parentheses i with hat on top plus j with hat on top minus 2 k with hat on top close parentheses equals j with hat on top plus 3 k with hat on top
N o w space c o n s i d e r space stack A B with rightwards arrow on top cross times stack A C with rightwards arrow on top colon space space
n with rightwards arrow on top equals stack A B with rightwards arrow on top cross times stack A C with rightwards arrow on top equals open vertical bar table row cell table row cell i with hat on top end cell cell j with hat on top end cell cell k with hat on top end cell row 1 cell negative 2 end cell 3 row 0 1 3 end table end cell end table close vertical bar
n with rightwards arrow on top equals i with hat on top open parentheses negative 6 minus 3 close parentheses minus 3 j with hat on top plus k with hat on top equals negative 9 i with hat on top minus 3 j with hat on top plus k with hat on top
S o comma space t h e space e q u a t i o n space o f space t h e space r e q u i r e d space p l a n e space i s space
open parentheses r with rightwards arrow on top minus a with rightwards arrow on top close parentheses times n with rightwards arrow on top equals 0
rightwards double arrow open parentheses r with rightwards arrow on top times n with rightwards arrow on top close parentheses equals open parentheses a with rightwards arrow on top times n with rightwards arrow on top close parentheses
rightwards double arrow open parentheses r with rightwards arrow on top times open parentheses negative 9 i with hat on top minus 3 j with hat on top plus k with hat on top close parentheses close parentheses equals open parentheses i with hat on top plus j with hat on top minus 2 k with hat on top close parentheses times open parentheses negative 9 i with hat on top minus 3 j with hat on top plus k with hat on top close parentheses
rightwards double arrow r with rightwards arrow on top times open parentheses 9 i with hat on top plus 3 j with hat on top minus k with hat on top close parentheses equals 14
A l s o comma space f i n d space t h e space c o o r d i n a t e s space o f space t h e space p o i n t space o f space i n t e r s e c t i o n space o f space t h i s space p l a n e space a n d space
t h e space l i n e space r with rightwards arrow on top equals 3 i with hat on top minus j with hat on top minus k with hat on top plus lambda open parentheses 2 i with hat on top minus 2 j with hat on top plus k with hat on top close parentheses
A n y space p o i n t space o n space t h e space l i n e space r with rightwards arrow on top equals 3 i with hat on top minus j with hat on top minus k with hat on top plus lambda open parentheses 2 i with hat on top minus 2 j with hat on top plus k with hat on top close parentheses space i s space o f space t h e space f o r m comma
open parentheses 3 plus 2 lambda comma space minus 1 minus 2 lambda comma space minus 1 plus lambda close parentheses
I f space t h e space p o i n t space P open parentheses 3 plus 2 lambda comma space minus 1 minus 2 lambda comma space minus 1 plus lambda close parentheses space l i e s space i n space t h e space p l a n e comma space
r with rightwards arrow on top times open parentheses 9 i with hat on top plus 3 j with hat on top minus k with hat on top close parentheses equals 14 comma space w e space h a v e comma
9 open parentheses 3 plus 2 lambda close parentheses minus 3 open parentheses 1 plus 2 lambda close parentheses minus open parentheses negative 1 plus lambda close parentheses equals 14
rightwards double arrow 27 plus 18 lambda minus 3 minus 6 lambda plus 1 minus lambda equals 14
rightwards double arrow 11 lambda equals negative 11
rightwards double arrow lambda equals negative 1
T h u s comma space t h e space r e q u i r e d space p o i n t space o f space i n t e r s e c t i o n space i s space
P open parentheses 3 plus 2 lambda comma space minus 1 minus 2 lambda comma space minus 1 plus lambda close parentheses
rightwards double arrow P open parentheses 3 plus 2 open parentheses negative 1 close parentheses comma space minus 1 minus 2 open parentheses negative 1 close parentheses comma space minus 1 plus open parentheses negative 1 close parentheses close parentheses
rightwards double arrow P open parentheses 1 comma space 1 comma space minus 2 close parentheses

space

Question 12

Solution 12

 

Question 13

Find the equation of a plane which passes through the point (3, 2, 0) and contains the line

  

Solution 13

  

Chapter 29 - The plane Excercise Ex. 29.14

Question 1

Find space the space shortest space distance space between space the space lines space fraction numerator straight x minus 2 over denominator negative 1 end fraction equals fraction numerator straight y minus 5 over denominator 2 end fraction equals fraction numerator straight z minus 0 over denominator 3 end fraction space and space
fraction numerator straight x minus 0 over denominator 2 end fraction equals fraction numerator straight y plus 1 over denominator negative 1 end fraction equals fraction numerator straight z minus 1 over denominator 2 end fraction.

Solution 1

Question 2

Find space the space shortest space distance space between space the space lines space fraction numerator straight x plus 1 over denominator 7 end fraction equals fraction numerator straight y plus 1 over denominator negative 6 end fraction equals fraction numerator straight z plus 1 over denominator 1 end fraction space and space
fraction numerator straight x minus 3 over denominator 1 end fraction equals fraction numerator straight y minus 5 over denominator negative 2 end fraction equals fraction numerator straight z minus 7 over denominator 1 end fraction.

Solution 2

straight l subscript 1 colon fraction numerator straight x plus 1 over denominator 7 end fraction equals fraction numerator straight y plus 1 over denominator negative 6 end fraction equals fraction numerator straight z plus 1 over denominator 1 end fraction
straight l subscript 2 colon fraction numerator straight x minus 3 over denominator 1 end fraction equals fraction numerator straight y minus 5 over denominator negative 2 end fraction equals fraction numerator straight z minus 7 over denominator 1 end fraction
Let space the space equation space of space the space plane space containing space straight l subscript 1 space be space straight a left parenthesis straight x plus 1 right parenthesis plus straight b left parenthesis straight y plus 1 right parenthesis plus straight c left parenthesis straight z plus 1 right parenthesis equals 0
Plane space is space parallel space to space straight l subscript 1 colon space 7 straight a minus 6 straight b plus straight c equals 0...... left parenthesis straight i right parenthesis
Plane space is space parallel space to space straight l subscript 2 colon space straight a minus 2 straight b plus straight c equals 0........ left parenthesis ii right parenthesis
Solving space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma
fraction numerator straight a over denominator negative 6 plus 2 end fraction equals fraction numerator straight b over denominator 1 minus 7 end fraction equals fraction numerator straight c over denominator negative 14 plus 6 end fraction
fraction numerator straight a over denominator negative 4 end fraction equals fraction numerator straight b over denominator negative 6 end fraction equals fraction numerator straight c over denominator negative 8 end fraction
therefore Equation space of space the space plane space is space minus 4 left parenthesis straight x plus 1 right parenthesis minus 6 left parenthesis straight y plus 1 right parenthesis minus 8 left parenthesis straight z plus 1 right parenthesis equals 0
4 left parenthesis straight x plus 1 right parenthesis plus 6 left parenthesis straight y plus 1 right parenthesis plus 8 left parenthesis straight z plus 1 right parenthesis equals 0 space is space the space equation space of space the space plane.

Question 3

Find the shortest distance between the lines

  

Solution 3

  

Chapter 29 - The plane Excercise Ex. 29.15

Question 1

Solution 1

Question 2

Solution 2

Question 3

Hence or otherwise deduce the length of the perpendicular.

Solution 3

Question 4

Solution 4

Question 5

 

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

F i n d space t h e space d i r e c t i o n space cos i n e s space o f space t h e space u n i t space v e c t o r space p e r p e n d i c u l a r space t o space t h e space p l a n e
r with rightwards arrow on top times open parentheses 6 i with hat on top minus 3 j with hat on top minus 2 k with hat on top close parentheses plus 1 equals 0 space p a s sin g space t h r o u g h space t h e space o r i g i n.

Solution 12

G i v e n space e q u a t i o n space o f space t h e space p l a n e space r with rightwards arrow on top times open parentheses 6 i with hat on top minus 3 j with hat on top minus 2 k with hat on top close parentheses plus 1 equals 0
T h u s comma space t h e space d i r e c t i o n space r a t i o s space n o r m a l space t o space t h e space p l a n e space a r e space 6 comma space minus 3 space a n d space minus 2
H e n c e space t h e space d i r e c t i o n space cos i n e s space t o space t h e space n o r m a l space t o space t h e space p l a n e space a r e
fraction numerator 6 over denominator square root of 6 squared plus open parentheses negative 3 close parentheses squared plus open parentheses negative 2 close parentheses squared end root end fraction comma fraction numerator negative 3 over denominator square root of 6 squared plus open parentheses negative 3 close parentheses squared plus open parentheses negative 2 close parentheses squared end root end fraction comma fraction numerator negative 2 over denominator square root of 6 squared plus open parentheses negative 3 close parentheses squared plus open parentheses negative 2 close parentheses squared end root end fraction
equals 6 over 7 comma fraction numerator negative 3 over denominator 7 end fraction comma fraction numerator negative 2 over denominator 7 end fraction
equals fraction numerator negative 6 over denominator 7 end fraction comma 3 over 7 comma 2 over 7
T h e space d i r e c t i o n space cos i n e s space o f space t h e space u n i t space v e c t o r space p e r p e n d i c u l a r space t o space t h e space p l a n e space a r e
s a m e space a s space t h e space d i r e c t i o n space cos i n e s space o f space t h e space n o r m a l space t o space t h e space p l a n e.
T h u s comma space t h e space d i r e c t i o n space cos i n e s space o f space t h e space u n i t space v e c t o r space p e r p e n d i c u l a r space t o space t h e space p l a n e
a r e colon space fraction numerator negative 6 over denominator 7 end fraction comma 3 over 7 comma 2 over 7

Question 13

F i n d space t h e space c o o r d i n a t e s space o f space t h e space f o o t space o f space t h e space p e r p e n d i c u l a r space d r a w n space f r o m space t h e space o r i g i n
t o space t h e space p l a n e space 2 x minus 3 y plus 4 z minus 6 equals 0

Solution 13

C o n s i d e r space t h e space g i v e n space e q u a t i o n space o f space t h e space p l a n e space 2 x minus 3 y plus 4 z minus 6 equals 0
T h e space d i r e c t i o n space r a t i o s space o f space t h e space n o r m a l space t o space t h e space p l a n e space a r e space 2 comma space minus 3 space a n d space 4
T h u s comma space t h e space d i r e c t i o space r a t i o s space o f space t h e space l i n e space p e r p e n d i c u l a r space t o space t h e space p l a n e space a r e space 2 comma space minus 3 space a n d space 4.
T h e space e q u a t i o n space o f space t h e space l i n e space p a s sin g space open parentheses x subscript 1 comma y subscript 1 comma z subscript 1 close parentheses space h a v i n g space d i r e c t i o n space r a t i o s space a comma b space a n d space c space i s
fraction numerator x minus x subscript 1 over denominator a end fraction equals fraction numerator y minus y subscript 1 over denominator b end fraction equals fraction numerator z minus z subscript 1 over denominator c end fraction
T h u s comma space t h e space e q u a t i o n space o f space t h e space l i n e space p a s sin g space t h r o u g h space t h e space o r i g i n space w i t h
d i r e c t i o n space r a t i o s space 2 comma space minus 3 space a n d space 4 space i s
fraction numerator x minus 0 over denominator 2 end fraction equals fraction numerator y minus 0 over denominator negative 3 end fraction equals fraction numerator z minus 0 over denominator 4 end fraction
rightwards double arrow x over 2 equals fraction numerator y over denominator negative 3 end fraction equals z over 4 equals r comma space w h e r e space r space i s space s o m e space c o n s tan t
A n y space p o i n t space o n space t h e space l i n e space i s space o f space t h e space f o r m space 2 r comma space minus 3 r space a n d space 4 r
I f space t h e space p o i n t space P open parentheses 2 r comma space minus 3 r comma space 4 r close parentheses space l i e s space o n space t h e space p l a n e space 2 x minus 3 y plus 4 z minus 6 equals 0 comma
i t space s h o u l d space s a t i s f i e s space t h e space e q u a t i o n comma space space 2 x minus 3 y plus 4 z minus 6 equals 0
T h u s comma space w e space h a v e comma
space 2 open parentheses 2 r close parentheses minus 3 open parentheses negative 3 r close parentheses plus 4 open parentheses 4 r close parentheses minus 6 equals 0
rightwards double arrow 4 r plus 9 r plus 16 r minus 6 equals 0
rightwards double arrow 29 r equals 6
rightwards double arrow r equals 6 over 29
T h u s comma space t h e space c o o r d i n a t e s space o f space t h e space p o i n t space o f space i n t e r s e c t i o n space o f space t h e space
p e r p e n d i c u l a r space f r o m space t h e space o r i g i n space a n d space t h e space p l a n e space a r e colon
P open parentheses 2 cross times 6 over 29 comma space minus 3 cross times 6 over 29 comma space 4 cross times 6 over 29 close parentheses equals P open parentheses 12 over 29 comma space 18 over 29 comma space 24 over 29 close parentheses

Question 14

Find the length and the foot of perpendicular from the point (1, 3/2, 2) to the plane 2x - 2y + 4z + 5 = 0.

Solution 14

  

Chapter 29 - The plane Excercise MCQ

Question 1

The plane 2x - (1 + λ)y + 3λz = 0 passes through the intersection of the planes

  1. 2x - y = 0 and y - 3z = 0
  2. 2x + 3z = 0 and y = 0
  3. 2x - y + 3z = 0 and y - 3z = 0
  4. none of these
Solution 1

Correct option: (a)

  

 

Question 2

The acute angle between the planes 2x - y + z = 6 and x + y + 2z = 3 is

  1. 45°
  2. 60°
  3. 30°
  4. 75°
Solution 2

Correct option: (b)

  

 

Question 3

The equation of the plane through the intersection of the planes x + 2y + 3z = 4 and 2x + y - z = -5 and perpendicular to the plane 5x + 3y + 6z + 8 = 0 is

 

  1. 7x - 2y + 3z + 81 = 0
  2. 23x + 14y - 9z + 48 = 0
  3. 51x - 15y - 50z + 173 = 0
  4. none of these
Solution 3

Correct option: (d)

  

 

Question 4

The distance between the planes 2x + 2y - z + 2 = 0 and 4x + 4y - 2z + 5 = 0 is

 

Solution 4

Correct option: (c)

  

 

Question 5

The image of the point (1, 3, 4) in the plane 2x - y + z + 3 = 0 is

 

  1. (3, 5, 2)
  2. (-3, 5, 2)
  3. (3, 5, -2)
  4. (3, -5, 2)
Solution 5

Correct option: (b)

  

Question 6

  1. 8x + y - 5z - 7 = 0
  2. 8x + y + 5z - 7 = 0
  3. 8x - y - 5z - 7 = 0
  4. none of these
Solution 6

Correct option: (d)

 

NOTE: Answer not matching with back answer.

Question 7

Solution 7

Correct option: (a)

  

Question 8

Solution 8

Correct option: (b)

  

Question 9

  1. x - 5y + 3z = 7
  2. x - 5y + 3z = -7
  3. x + 5y + 3z = 7
  4. x + 5y + 3z = -7
Solution 9

Correct option: (a)

  

Question 10

Solution 10

Correct option: (a)

  

Question 11

  1. 1
  2. 2
  3. 3
  4. none of these
Solution 11

Correct option: (c)

  

Question 12

  1. 1
  2. 2
  3. 3 
  4. none of these
Solution 12

Correct option: (c)

  

Question 13

Solution 13

Correct option: (a)

  

Question 14

Solution 14

Correct option: (c)

  

Question 15

The equation of the plane parallel to the lines x - 1 = 2y - 5 = 2z and 3x = 4y - 11 = 3z - 4 and passing through the point (2, 3, 3) is

  1. x - 4y + 2z + 4 = 0
  2. x + 4y + 2z + 4 = 0
  3. x - 4y + 2z - 4 = 0
  4. none of these
Solution 15

Correct option: (a)

  

Question 16

  1. 9
  2. 13
  3. 17
  4. none of these
Solution 16

Correct option: (b)

  

Question 17

The equation of the plane through the intersection of the planes ax + by + cz + d = 0 and lx + my + nz + p = 0 and parallel to the line y = 0, z = 0

  1. (bl - am)y + (cl - an)z + dl - ap = 0
  2. (am - bl)x + (mc - bn)z + md - bp = 0
  3. (na - cl)x + (bn - cm)y + nd - cp = 0
  4. none of these
Solution 17

Correct option: (a)

  

Question 18

The equation of the plane which cuts equal intercepts of unit length on the coordinate axes is

  1. x + y + z = 1
  2. x + y + z = 0
  3. x + y - z = 1
  4. x + y + z = 2
Solution 18

Correct option: (a)

  

Chapter 29 - The plane Excercise Ex. 29VSAQ

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Find the length of the perpendicular drawn from the origin to the plane 2x - 3y + 6z + 21 = 0.

Solution 19

begin mathsize 12px style table attributes columnalign left end attributes row cell text The   length   of   the   perpendicular   drawn   from   end text end cell row cell text the   origin   to   the   plane   end text 2 straight x minus 3 straight y plus 6 straight z plus 21 equals 0 text   is end text end cell row cell fraction numerator text 2 end text left parenthesis text 0 end text right parenthesis minus 3 left parenthesis 0 right parenthesis plus 6 left parenthesis 0 right parenthesis plus 21 over denominator square root of 2 squared plus 3 squared plus 6 squared end root end fraction equals fraction numerator 21 over denominator square root of 49 end fraction equals 21 over 7 equals 3 end cell end table end style

Question 20

begin mathsize 12px style table attributes columnalign left end attributes row cell text Write   the   vector   equation   of   the   line   passing   through   end text end cell row cell text the   point   end text left parenthesis text 1 ,- 2 ,- 3 end text right parenthesis text and   normal   to   the   plane   end text stack text r end text with rightwards arrow on top. open parentheses 2 straight i with hat on top plus straight j with hat on top plus 2 straight k with hat on top close parentheses equals 5. end cell end table end style

Solution 20

begin mathsize 12px style table attributes columnalign left end attributes row cell text Position   vector   of   the   point   is   end text straight i with hat on top minus 2 straight j with hat on top minus 3 straight k with hat on top end cell row cell text A   vector   perpendicular   to   the   plane   is   end text 2 straight i with hat on top plus straight j with hat on top plus 2 straight k with hat on top end cell row cell text Hence ,  equation   of   the   line   is end text end cell row cell straight i with hat on top minus 2 straight j with hat on top minus 3 straight k with hat on top plus straight lambda open parentheses 2 straight i with hat on top plus straight j with hat on top plus 2 straight k with hat on top close parentheses end cell end table end style

Question 21

begin mathsize 12px style table attributes columnalign left end attributes row cell text Write   the   vector   equation   of   the   plane ,  passing   through   end text end cell row cell text the   point   end text left parenthesis text a , b , c end text right parenthesis text and   parallel   to   the   plane   end text stack text r end text with rightwards arrow on top. open parentheses straight i with hat on top plus straight j with hat on top plus straight k with hat on top close parentheses equals 2. end cell end table end style

Solution 21

begin mathsize 12px style table attributes columnalign left end attributes row cell text The   vector   equation   of   a   plane   passing   through   end text end cell row cell text a   point   having   position   vector   end text stack text A end text with rightwards arrow on top text   and   normal   to   end text stack text n end text with rightwards arrow on top text   end text end cell row cell text is   end text left square bracket stack text r end text with rightwards arrow on top minus stack text A end text with rightwards arrow on top right square bracket times stack text n end text with rightwards arrow on top equals 0 end cell row cell text Here ,  end text stack text n end text with rightwards arrow on top equals straight i with hat on top plus straight j with hat on top plus straight k with hat on top text   end text open parentheses because text It   is   parallel   to   the   plane   end text stack text r end text with rightwards arrow on top times equals open parentheses straight i with hat on top plus straight j with hat on top plus straight k with hat on top close parentheses 2 close parentheses end cell row cell therefore text The   required   equation   of   the   plane ,  end text end cell row cell text is   end text open square brackets stack text r end text with rightwards arrow on top minus open parentheses straight a straight i with hat on top plus straight b straight j with hat on top plus straight c straight k with hat on top close parentheses close square brackets times open parentheses straight i with hat on top plus straight j with hat on top plus straight k with hat on top close parentheses equals 0 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times open parentheses straight i with hat on top plus straight j with hat on top plus straight k with hat on top close parentheses equals straight a plus straight b plus straight c end cell end table end style