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Vector Algebra

Question
CBSEENMA12034101
Wired Faculty App

Find the projection of the vector straight i with hat on top space plus space 3 space straight j with hat on top space plus space 7 space straight k with hat on top on the vector 7 space straight i with hat on top space minus space straight j with hat on top space plus space 8 space straight k with hat on top.

Solution

Let straight a with rightwards arrow on top space equals space straight i with hat on top plus 3 straight j with hat on top plus 7 straight k with hat on top comma space space straight b with rightwards arrow on top space equals space 7 straight i with hat on top minus straight j with hat on top plus 8 straight k with hat on top
therefore space space space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space left parenthesis 1 right parenthesis thin space left parenthesis 7 right parenthesis space plus space left parenthesis 3 right parenthesis thin space left parenthesis negative 1 right parenthesis space plus space left parenthesis 7 right parenthesis thin space left parenthesis 8 right parenthesis space equals space 7 minus 3 space plus space 56 space equals space 60 space space space space space space space space space space space straight b space equals space square root of 49 plus 1 plus 64 end root space equals space square root of 114 Required space projection space equals space fraction numerator straight a with rightwards arrow on top. space straight b with rightwards arrow on top over denominator straight b end fraction space equals space fraction numerator 60 over denominator square root of 114 end fraction

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