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Class 12-commerce RD SHARMA Solutions Maths Chapter 23 - Algebra of Vectors

Algebra of Vectors Exercise Ex. 23.1

Solution 1(i)

Solution 1(ii)

Here, vector begin mathsize 12px style OP with rightwards arrow on top end style represents the displacement of 50 km, south-east.

Solution 1(iii)

Here, vector begin mathsize 12px style OP with rightwards arrow on top end style represents the displacement of 70 km, 40° north of west.

Solution 2

(i) 15 kg is a scalar quantity because it involves only

(ii) 20 kg weight is a vector quantity as it involves both magnitude and direction.

(iii) 45° is a scalar quantity as it involves only magnitude.

(iv) 10 metres south-east is a vector quantity as it involve direction.

(v) 50 m/s2 is a scalar quantity as it involves magnitude of acceleration.

Solution 3

Solution 4

Solution 5

Algebra of Vectors Exercise Ex. 23.2

Solution 1

Solution 2

      

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Algebra of Vectors Exercise Ex. 23.3

Solution 1

 

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

L e t space A B C space b e space a space t r i a n g l e. L e t space t h e space p o s i t i o n space v e c t o r s space o f space A comma space B space a n d space C space w i t h space r e s p e c t space t o space s o m e space o r i g i n comma space O space b e a with rightwards arrow on top comma space b with rightwards arrow on top space a n d space c with rightwards arrow on top space r e s p e c t i v e l y. L e t space D space b e space t h e space p o i n t space o n space B C space w h e r e space t h e space b i s e c t o r space o f space angle A space m e e t s. L e t space d with rightwards arrow on top space p o s i t i o n space v e c t o r space o f space D space w h i c h space d i v i d e s space B C space i n t e r n a l l y space i n space t h e space r a t i o space beta space a n d space gamma comma w h e r e space beta equals open vertical bar stack A C with rightwards arrow on top close vertical bar space a n d space gamma equals open vertical bar stack A B with rightwards arrow on top close vertical bar T h u s comma space beta equals open vertical bar c with rightwards arrow on top minus a with rightwards arrow on top close vertical bar space a n d space gamma equals open vertical bar b with rightwards arrow on top minus a with rightwards arrow on top close vertical bar T h u s comma space b y space s e c t i o n space f o r m u l a comma space t h e space p o s i t i o n space v e c t o r space o f space D space i s space g i v e n space b y stack O D with rightwards arrow on top equals fraction numerator beta b with rightwards arrow on top plus gamma c with rightwards arrow on top over denominator beta plus gamma end fraction L e t space alpha equals open vertical bar b with rightwards arrow on top minus c with rightwards arrow on top close vertical bar I n c e n t r e space i s space t h e space c o n c u r r e n t space p o i n t space o f space a n g l e space b i s e c t o r s. T h u s comma space I n c e n t r e space d i v i d e s space t h e space l i n e space A D space i n space t h e space r a t i o space alpha : beta plus gamma T h u s comma space t h e space p o s i t i o n space v e c t o r space o f space i n c e n t r e space i s e q u a l space t o space fraction numerator alpha a with rightwards arrow on top plus fraction numerator beta b with rightwards arrow on top plus gamma c with rightwards arrow on top over denominator open parentheses beta plus gamma close parentheses end fraction cross times open parentheses beta plus gamma close parentheses over denominator alpha plus beta plus gamma end fraction equals fraction numerator alpha a with rightwards arrow on top plus beta b with rightwards arrow on top plus gamma c with rightwards arrow on top over denominator alpha plus beta plus gamma end fraction  

 


Algebra of Vectors Exercise Ex. 23.4

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5


Solution 6

 

 

 

 

 

Algebra of Vectors Exercise Ex. 23.5

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

rightwards double arrow b with hat on top equals 1 half i with hat on top plus fraction numerator square root of 3 over denominator 2 end fraction j with hat on top

Solution 12

Algebra of Vectors Exercise Ex. 23.6

Solution 1

Solution 2

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Solution 3

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

G i v e n space t h a t space a with rightwards arrow on top equals 2 stack i space with hat on top plus 3 j with hat on top minus k with hat on top space a n d b with rightwards arrow on top equals i with hat on top minus 2 j with hat on top plus k with hat on top T h u s comma space F i n d space a space v e c t o r space o f space m a g n i t u d e space o f space 5 space u n i t s space p a r a l l e l space t o space t h e space r e s u l tan t space o f space t h e space v e c t o r s space a with rightwards arrow on top plus b with rightwards arrow on top equals 2 i with hat on top plus 3 j with hat on top minus k with hat on top plus space i with hat on top minus 2 j with hat on top plus k with hat on top rightwards double arrow a with rightwards arrow on top plus b with rightwards arrow on top equals 3 i with hat on top plus j with hat on top rightwards double arrow open vertical bar a with rightwards arrow on top plus b with rightwards arrow on top close vertical bar equals square root of 9 plus 1 end root equals square root of 10 T h u s comma space t h e space u n i t space v e c t o r space a l o n g space t h e space r e s u l tan t space v e c t o r space a with rightwards arrow on top plus b with rightwards arrow on top space i s space fraction numerator 3 i with hat on top plus j with hat on top over denominator square root of 10 end fraction T h e space v e c t o r space o f space m a g n i t u d e space o f space 5 space u n i t s space p a r a l l e l space t o space t h e space r e s u l tan t v e c t o r equals fraction numerator 3 i with hat on top plus j with hat on top over denominator square root of 10 end fraction cross times 5 equals square root of 5 over 2 end root open parentheses 3 i with hat on top plus j with hat on top close parentheses

Solution 19

 

Algebra of Vectors Exercise Ex. 23.7

Solution 1

Solution 2 (i)

Solution 2 (ii)

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

W e space h a v e stack A P with rightwards arrow on top equals P o s i t i o n space v e c t o r space o f space P minus P o s i t i o n space v e c t o r space o f space A rightwards double arrow stack A P with rightwards arrow on top equals i with hat on top plus 2 j with hat on top plus 3 k with hat on top minus open parentheses minus 2 i with hat on top plus 3 j with hat on top plus 5 k with hat on top close parentheses equals 3 i with hat on top minus j with hat on top minus 2 k with hat on top stack P B with rightwards arrow on top equals P o s i t i o n space v e c t o r space o f space B minus P o s i t i o n space v e c t o r space o f space P rightwards double arrow stack P B with rightwards arrow on top equals 7 i with hat on top minus k with hat on top minus open parentheses i with hat on top plus 2 j with hat on top plus 3 k with hat on top close parentheses equals 6 i with hat on top minus 2 j with hat on top minus 4 k with hat on top C l e a r l y comma space stack P B with rightwards arrow on top equals 2 stack A P with rightwards arrow on top s o space v e c t o r s space stack A P with rightwards arrow on top space a n d space stack P B with rightwards arrow on top space a r e space c o l l i n e a r. B u t space P space i s space a space p o i n t space c o m m o n space t o space stack A P with rightwards arrow on top space a n d space stack P B with rightwards arrow on top. space H e n c e space P comma space A comma space B space a r e space c o l l i n e a r space p o i n t s. S i m i l a r l y comma space stack C P with rightwards arrow on top equals i with hat on top plus 2 j with hat on top plus 3 k with hat on top minus open parentheses minus 3 i with hat on top minus 2 j with hat on top minus 5 k with hat on top close parentheses equals 4 i with hat on top plus 4 j with hat on top plus 8 k with hat on top a n d space stack P D with rightwards arrow on top equals 3 i with hat on top plus 4 j with hat on top plus 7 k with hat on top minus open parentheses i with hat on top plus 2 j with hat on top plus 3 k with hat on top close parentheses equals 2 i with hat on top plus 2 j with hat on top plus 4 k with hat on top S o space v e c t o r s space stack C P with rightwards arrow on top space a n d space stack P D with rightwards arrow on top space a r e space c o l l i n e a r. B u t space P space i s space a space c o m m o n space p o i n t space t o space stack C P with rightwards arrow on top space a n d space stack C D with rightwards arrow on top. H e n c e comma space C comma space P comma space D space a r e space c o l l i n e a r space p o i n t s. T h u s comma space A comma space B comma space C comma space D space a n d space P space a r e space p o i n t s space s u c h space t h a t space A comma space P comma space B space a n d space C comma space P comma space D space a r e space t w o space s e t s space o f space c o l l i n e a r space p o i n t s. space H e n c e space A B space a n d space C D space i n t e r s e c t space a t space t h e p o i n t space P

Solution 13

 

Algebra of Vectors Exercise Ex. 23.8

Solution 1

Solution 2 (i)

Solution 2 (ii)

Solution 2 (iii)

Solution 2 (iv)

Solution 3 (i)

Solution 3 (ii)

Solution 4

 

Solution 5 (i)

Solution 5 (ii)

Solution 6 (i)

Solution 6 (ii)

Solution 7 (i)

Solution 7 (ii)

Solution 8

Solution 9

Solution 10

Algebra of Vectors Exercise Ex. 23.9

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7 (i)

Solution 7 (ii)

Solution 7 (iii)

Solution 8

Solution 9

Solution 10

Solution 11

 

Solution 12

 

Algebra of Vectors Exercise MCQ

Solution 1

Correct option:(a)

 

Solution 2

Correct option:(c)

 

 

Solution 3

Correct option: (c)

 

 

 

 

Solution 4

Correct option: (d)

 

 

Solution 5

Correct option: (b)

 

Solution 6

Correct option: (b)

 

Solution 7

Correct option: (b)

 

Solution 8

Correct option: (c)

 

 

Solution 9

Correct option: (b)

 

Solution 10

Correct option: (b)

 

Solution 11

Correct option: (d)

 

Solution 12

Correct option: (a)

NOTE: Answer not matching with back answer.

Solution 13

Correct option: (d)

 

 

 

Solution 14

Correct option: (a)

 

Solution 15

Correct option: (a)

 

Solution 16

Correct option: (c)

 

 

 

 

Solution 17

Correct option: (d)

 

Solution 18

Correct option: (d)

 

Solution 19

Correct option: (c)

 

Algebra of Vectors Exercise Ex. 23VSAQ

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

I n space a n space e q u i l a t e r a l space t r i a n g l e comma space t h e space c e n t r o i d comma space o r t h o c e n t r e comma space i n c e n t r e space a n d space c i r c u m c e n t r e c o i n c i d e. S i n c e space t h e space o r t h o c e n t r e space i s space a t space t h e space o r i g i n comma space c e n t r o i d space i s space a l s o space a t space t h e space o r i g i n. P o s i t i o n space v e c t o r space o f space C e n t r o i d space i s space fraction numerator a with rightwards arrow on top plus b with rightwards arrow on top plus c with rightwards arrow on top over denominator 3 end fraction equals 0 with rightwards arrow on top T h u s comma space w e space h a v e comma space a with rightwards arrow on top plus b with rightwards arrow on top plus c with rightwards arrow on top space equals space 0 with rightwards arrow on top

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

L e t space u s space c o n s i d e r space t h e space t w o space v e c t o r s space a with rightwards arrow on top space a n d space b with rightwards arrow on top space s u c h space t h a t space a with rightwards arrow on top equals 3 i with hat on top minus j with hat on top plus 3 k with hat on top space a n d b with rightwards arrow on top equals i with hat on top plus 3 j with hat on top minus 3 k with hat on top T h e space m a g n i t u d e space o f space a space i s space open vertical bar a with rightwards arrow on top close vertical bar equals square root of 3 squared plus open parentheses minus 1 close parentheses squared plus 3 squared end root equals square root of 19 T h e space m a g n i t u d e space o f space b space i s space open vertical bar b with rightwards arrow on top close vertical bar equals square root of 1 squared plus 3 squared plus open parentheses minus 3 close parentheses squared end root equals square root of 19 I t space i s space c l e a r space t h a t space space open vertical bar a with rightwards arrow on top close vertical bar equals open vertical bar b with rightwards arrow on top close vertical bar

Solution 36

L e t space u s space c o n s i d e r space t h e space t w o space v e c t o r s space a with rightwards arrow on top space a n d space b with rightwards arrow on top space s u c h space t h a t space a with rightwards arrow on top equals i with hat on top minus j with hat on top plus 3 k with hat on top space a n d b with rightwards arrow on top equals 2 i with hat on top minus 2 j with hat on top plus 6 k with hat on top I t space i s space c l e a r space t h a t space space b with rightwards arrow on top equals 2 i with hat on top minus 2 j with hat on top plus 6 k with hat on top equals 2 open parentheses i with hat on top minus j with hat on top plus 3 k with hat on top space close parentheses equals k a with rightwards arrow on top T h u s comma space a with rightwards arrow on top space i s space p a r a l l e l space t o space b with rightwards arrow on top space a n d space h e n c e space i n space t h e space s a m e space d i r e c t i o n.

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

begin mathsize 12px style table attributes columnalign left end attributes row cell straight a with rightwards arrow on top text  =  end text straight x straight i with hat on top plus 2 straight j with hat on top minus straight z straight k with hat on top text   and   end text straight b with rightwards arrow on top equals text 3 end text straight i with hat on top minus straight y straight j with hat on top plus straight k with hat on top text   are   two   equal   vectors. end text end cell row cell text For   equal   vectors ,  the   components   are   equal. end text end cell row cell text Hence ,  end text straight x equals 3 comma text   end text 2 equals negative straight y comma text   end text minus straight z equals 1 end cell row cell straight x plus straight y plus straight z equals 3 minus 2 minus 1 equals 0 end cell end table end style

Solution 45

begin mathsize 12px style table attributes columnalign left end attributes row cell text   end text straight a with rightwards arrow on top text  =  2 end text straight i with hat on top plus 2 straight j with hat on top minus 5 straight k with hat on top text   and   end text straight b with rightwards arrow on top equals text 2 end text straight i with hat on top plus straight j with hat on top minus 7 straight k with hat on top end cell row cell straight a with rightwards arrow on top plus straight b with rightwards arrow on top equals left parenthesis 2 plus 2 right parenthesis straight i with hat on top plus left parenthesis 2 plus 1 right parenthesis straight j with hat on top minus left parenthesis 5 plus 7 right parenthesis straight k with hat on top end cell row cell equals 4 straight i with hat on top plus 3 straight j with hat on top minus 12 straight k with hat on top end cell row cell text Unit   vector   in   the   direction   of   end text straight a with rightwards arrow on top plus straight b with rightwards arrow on top end cell row cell equals fraction numerator straight a with rightwards arrow on top plus straight b with rightwards arrow on top over denominator vertical line straight a with rightwards arrow on top plus straight b with rightwards arrow on top vertical line end fraction equals fraction numerator 4 straight i with hat on top plus 3 straight j with hat on top minus 12 straight k with hat on top over denominator square root of 4 squared plus 3 squared plus 12 squared end root end fraction equals fraction numerator 4 straight i with hat on top plus 3 straight j with hat on top minus 12 straight k with hat on top over denominator square root of 169 end fraction equals 4 over 13 straight i with hat on top plus 3 over 13 straight j with hat on top minus 12 over 13 straight k with hat on top end cell end table end style

Solution 46

begin mathsize 12px style table attributes columnalign left end attributes row cell text If   two   vectors   end text straight a with rightwards arrow on top text   and   end text straight b with rightwards arrow on top text   are   parallel ,  then   end text straight a with rightwards arrow on top text  =  end text straight lambda straight b with rightwards arrow on top. end cell row cell text Here   end text straight lambda text   is   a   constant. end text end cell row cell text 3 end text straight i with hat on top plus 2 straight j with hat on top plus 9 straight k with hat on top text   and   end text straight i with hat on top minus 2 straight p straight j with hat on top plus 3 straight k with hat on top text   are   parallel. end text end cell row cell text Hence ,  3 end text straight i with hat on top plus 2 straight j with hat on top plus 9 straight k with hat on top equals straight lambda open parentheses straight i with hat on top minus 2 straight p straight j with hat on top plus 3 straight k with hat on top close parentheses end cell row cell rightwards double arrow text 3 end text straight i with hat on top plus 2 straight j with hat on top plus 9 straight k with hat on top equals straight lambda straight i with hat on top minus 2 λp straight j with hat on top plus 3 straight lambda straight k with hat on top end cell row cell text Equating   the   components   we   get end text end cell row cell straight lambda equals 3 text   and   end text minus 2 λp equals 2 rightwards double arrow 6 straight p equals negative 2 rightwards double arrow straight p equals negative 1 third end cell end table end style

Solution 47

begin mathsize 12px style table attributes columnalign left end attributes row cell text Vector   end text straight a with rightwards arrow on top text   of   magnitude   5 end text square root of text 2 end text end root text   makes   an   angle   of   end text fraction numerator straight pi over denominator text 4 end text end fraction text   with   the   x-axis. end text end cell row cell text Hence ,  the   component   along   the   x-axis   is   end text 5 square root of 2 text cos end text fraction numerator straight pi over denominator text 4 end text end fraction text  =  end text 5 square root of 2 cross times fraction numerator text 1 end text over denominator square root of text 2 end text end root end fraction text = 5   end text end cell row cell text Vector   end text straight a with rightwards arrow on top text   of   magnitude   5 end text square root of text 2 end text end root text   makes   an   angle   of   end text fraction numerator straight pi over denominator text 2 end text end fraction text   with   the   y-axis. end text end cell row cell text Hence ,  the   component   along   the   y-axis   is   end text 5 square root of 2 text cos end text fraction numerator straight pi over denominator text 2 end text end fraction text  =  end text 5 square root of 2 cross times 0 text   =  0 end text end cell row cell text Vector   end text straight a with rightwards arrow on top text   of   magnitude   5 end text square root of text 2 end text end root text   makes   an   acute   angle   end text straight theta text   with   the   z-axis. end text end cell row cell text Hence ,  the   component   along   the   z-axis   is   end text 5 square root of 2 text cos end text straight theta. end cell row cell left parenthesis text 5 end text square root of text 2 end text end root right parenthesis squared equals 5 squared plus 0 squared plus left parenthesis 5 square root of 2 text cos end text straight theta right parenthesis squared equals 50 equals 25 plus 50 cos squared straight theta end cell row cell rightwards double arrow cos squared straight theta equals 25 over 50 equals 1 half end cell row cell rightwards double arrow cosθ equals fraction numerator 1 over denominator square root of 2 end fraction... left parenthesis because straight theta text   is   acute end text right parenthesis end cell row cell text Hence ,  end text straight a with rightwards arrow on top equals 5 straight i with hat on top plus 5 straight k with hat on top end cell end table end style

Solution 48

begin mathsize 12px style table attributes columnalign left end attributes row cell straight P text   and   end text straight Q text   are   the   points   with   co-ordinates  ( 1 , 3 , 0 )  and  ( 4 , 5 , 6 )  respectively. end text end cell row cell stack text PQ end text with rightwards arrow on top equals left parenthesis 4 minus 1 right parenthesis straight i with hat on top plus left parenthesis 5 minus 3 right parenthesis straight j with hat on top plus left parenthesis 6 minus 0 right parenthesis straight k with hat on top end cell row cell equals 3 straight i with hat on top plus 2 straight j with hat on top plus 6 straight k with hat on top end cell row cell text Unit   vector   in   the   direction   of   end text stack text PQ end text with rightwards arrow on top equals fraction numerator 3 straight i with hat on top plus 2 straight j with hat on top plus 6 straight k with hat on top over denominator square root of 3 squared plus 2 squared plus 6 squared end root end fraction equals fraction numerator 3 straight i with hat on top plus 2 straight j with hat on top plus 6 straight k with hat on top over denominator square root of 49 end fraction end cell row cell equals 3 over 7 straight i with hat on top plus 2 over 7 straight j with hat on top plus 6 over 7 straight k with hat on top end cell end table end style

Solution 49

begin mathsize 12px style table attributes columnalign left end attributes row cell text Unit   vector   in   the   direction   of   vector   2 end text straight i with hat on top minus 3 straight j with hat on top plus 6 straight k with hat on top end cell row cell equals fraction numerator text 2 end text straight i with hat on top minus 3 straight j with hat on top plus 6 straight k with hat on top over denominator square root of 2 squared plus 3 squared plus 6 squared end root end fraction equals fraction numerator text 2 end text straight i with hat on top minus 3 straight j with hat on top plus 6 straight k with hat on top over denominator 7 end fraction end cell row cell text Hence ,  a   vector   along   the   above   vector end text end cell row cell text which   has   magnitude   21   units end text equals 21 open parentheses fraction numerator text 2 end text straight i with hat on top minus 3 straight j with hat on top plus 6 straight k with hat on top over denominator 7 end fraction close parentheses end cell row cell equals 6 straight i with hat on top minus 9 straight j with hat on top plus 18 straight k with hat on top end cell end table end style

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