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Vector Algebra

Question
CBSEENMA12034062
Wired Faculty App

If straight a with rightwards arrow on top space equals space straight i with hat on top plus space 2 space straight j with hat on top space space minus space 3 space straight k with hat on top comma space space straight b with rightwards arrow on top space equals space 3 space straight i with hat on top space minus space straight j with hat on top space plus space 2 space straight k with hat on top comma space space show that open parentheses straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top close parentheses space and space left parenthesis straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top right parenthesis are perpendicular to each other. 

Solution
straight a 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 3 space straight k with hat on top comma space space straight b with rightwards arrow on top space equals space 3 space straight i with hat on top space minus space straight j with hat on top space plus space 2 space straight k with hat on top therefore space space space space space space space straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top space equals space 4 space straight i with hat on top space plus space straight j with hat on top space minus space straight k with hat on top comma space space space straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top space equals space minus 2 space straight i with hat on top space plus space 3 space straight j with hat on top space minus space 5 space straight k with hat on top
Now,    open parentheses straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top close parentheses space. space left parenthesis straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top right parenthesis space equals space 4 space left parenthesis negative 2 right parenthesis space plus space left parenthesis 1 right parenthesis thin space left parenthesis 3 right parenthesis thin space plus space left parenthesis negative 1 right parenthesis thin space left parenthesis negative 5 right parenthesis space space space
                                             equals negative 8 plus 3 plus 5 equals space 0
therefore space space straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top space space and space straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top are perpendicular to each other. 

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