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CBSE Class 12-science Answered

Let  rightwards arrow for A of be vector parallel to the line of intersection of  planes P1 and P2. Plane P1 is parallel to the
 
vectors 2j^ +3k^ and 4j^ - 3k^ and plane P2 is parallel to j^- k^ and 3i^ + 3j^ . Find the angle
 
between the vector rightwards arrow for A of and a given vector 2i^ + j^ - 2k^
Asked by ahuja8087 | 14 Feb, 2017, 07:22: PM
answered-by-expert Expert Answer
begin mathsize 16px style Plane space straight P 1 space is space parallel space to space the space vector space 2 straight i with hat on top plus 3 straight k with hat on top space space and space space 4 straight j with hat on top minus 3 straight k with hat on top space The space normal space of space straight P 1 equals open parentheses 2 straight i with hat on top plus 3 straight k with hat on top close parentheses space cross times space open parentheses 4 straight j with hat on top minus 3 straight k with hat on top close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 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 space space space space straight k with hat on top end cell row 2 0 cell space space space space 3 end cell row 0 4 cell negative 3 end cell end table close vertical bar space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals straight i with hat on top left parenthesis 0 minus 12 right parenthesis minus straight j with hat on top left parenthesis negative 6 minus 0 right parenthesis plus space straight k with hat on top left parenthesis 8 minus 0 right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals negative 12 straight i with hat on top plus 6 straight j with hat on top plus 8 space straight k with hat on top The space normal space of space straight P 2 equals open parentheses straight j with hat on top minus straight k with hat on top space close parentheses cross times space open parentheses 3 straight i with hat on top plus 3 straight j with hat on top close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 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 0 1 cell space space space space minus 1 end cell row 3 3 0 end table close vertical bar space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals straight i with hat on top left parenthesis 0 plus 3 right parenthesis minus straight j with hat on top left parenthesis 0 plus 3 right parenthesis plus space straight k with hat on top left parenthesis 0 minus 3 right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 3 straight i with hat on top minus 3 straight j with hat on top minus 3 space straight k with hat on top The space line space of space intersection space of space the space planes space will space be space equals open parentheses negative 12 straight i with hat on top plus 6 straight j with hat on top plus 8 space straight k with hat on top space close parentheses cross times open parentheses 3 straight i with hat on top minus 3 straight j with hat on top minus 3 space straight k with hat on top close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 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 space straight k with hat on top end cell row cell negative 12 end cell 6 cell space space space space 8 end cell row 3 cell negative 3 end cell cell negative 3 end cell end table close vertical bar space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals left parenthesis negative 18 plus 24 right parenthesis straight i with hat on top minus open parentheses 36 minus 24 space close parentheses straight j with hat on top plus straight k with hat on top open parentheses 36 minus 18 close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 6 straight i with hat on top minus 12 straight j with hat on top plus 18 space straight k with hat on top Now comma space angle space between space straight A with rightwards arrow on top equals 6 straight i with hat on top minus 12 straight j with hat on top plus 18 space straight k with hat on top space and space 2 straight i with hat on top plus straight j with hat on top minus 2 space straight k with hat on top space is space given space by space cos space straight theta equals fraction numerator left parenthesis 6 right parenthesis left parenthesis 2 right parenthesis plus left parenthesis negative 12 right parenthesis left parenthesis 1 right parenthesis plus left parenthesis 18 right parenthesis left parenthesis negative 2 right parenthesis over denominator square root of 6 squared plus left parenthesis negative 12 right parenthesis squared plus left parenthesis 18 right parenthesis squared end root space square root of 2 squared plus left parenthesis 1 right parenthesis squared plus left parenthesis negative 2 right parenthesis squared end root end fraction rightwards double arrow cos space straight theta equals fraction numerator 12 minus 12 minus 36 over denominator square root of 504 space square root of 9 end fraction rightwards double arrow cos space straight theta equals fraction numerator negative 36 over denominator square root of 4 cross times 126 end root space left parenthesis 3 right parenthesis end fraction rightwards double arrow cos space straight theta equals fraction numerator negative 36 over denominator 2 square root of 126 space left parenthesis 3 right parenthesis end fraction rightwards double arrow cos space straight theta equals fraction numerator negative 6 over denominator 3 square root of 14 space end fraction rightwards double arrow straight theta equals cos to the power of negative 1 end exponent open parentheses fraction numerator negative 2 over denominator square root of 14 space end fraction close parentheses comma space which space is space the space required space angle. end style
Answered by Rebecca Fernandes | 15 Feb, 2017, 09:54: AM

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CBSE 12-science - Maths
Q 48
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