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Vector Algebra

Question
CBSEENMA12034102
Wired Faculty App

Find the projection of 8 space straight i with hat on top space plus space straight j with hat on top in the direction of straight i with hat on top space plus space 2 space straight j with hat on top space minus space 2 space straight k with hat on top.

Solution

Let straight a with rightwards arrow on top space equals space 8 space straight i with hat on top space plus space straight j with hat on top comma space space space straight b with rightwards arrow on top space equals space straight i with hat on top space plus space 2 space straight j with hat on top space minus space 2 space 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 8 right parenthesis thin space left parenthesis 1 right parenthesis space plus space left parenthesis 1 right parenthesis thin space left parenthesis 2 right parenthesis space plus space left parenthesis 0 right parenthesis thin space left parenthesis negative 2 right parenthesis space equals space 8 plus 2 minus 0 space equals space 10 space space space space space space space space space space space straight b space equals space square root of 1 plus 4 plus 4 end root space equals square root of 9 space equals space 3
Required projection  = 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 10 over 3

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