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Vector Algebra

Question
CBSEENMA12034065
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If straight a with rightwards arrow on top space equals space 3 straight i with hat on top space plus space 2 straight j with hat on top space plus space 9 straight k with hat on top space and space straight b with rightwards arrow on top space equals space straight i with hat on top space plus space straight lambda straight j with hat on top space plus space 3 straight k with hat on top comma find the value of straight lambda so that straight a with rightwards arrow on top plus straight b with rightwards arrow on top is perpendicular to straight a with rightwards arrow on top minus straight b with rightwards arrow on top.

Solution

          straight a with rightwards arrow on top space equals space 3 straight i with hat on top space plus space 2 straight j with hat on top space plus space 9 straight k with hat on top comma space space space space straight b with rightwards arrow on top space equals space straight i with hat on top space plus space straight lambda straight j with hat on top space plus space 3 straight k with hat on top
therefore space space space straight a with rightwards arrow on top plus straight b with rightwards arrow on top space equals space 4 straight i with hat on top plus left parenthesis straight lambda plus 2 right parenthesis space straight j with hat on top plus 12 straight k with hat on top comma space space straight a with rightwards arrow on top minus straight b with rightwards arrow on top space equals space 2 space straight i with hat on top plus space left parenthesis 2 minus space straight lambda right parenthesis space straight j with hat on top space plus space 6 straight k with hat on top space
Since straight a with rightwards arrow on top plus straight b with rightwards arrow on top is perpendicular to straight a with rightwards arrow on top minus straight b with rightwards arrow on top
therefore space space space space space space space space space space space space left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis. space left parenthesis straight a with rightwards arrow on top space minus straight b with rightwards arrow on top right parenthesis space equals space 0 therefore space space space space space space left parenthesis 4 right parenthesis thin space left parenthesis 2 right parenthesis space plus space left parenthesis straight lambda plus 2 right parenthesis thin space left parenthesis 2 minus straight lambda right parenthesis space plus space left parenthesis 12 right parenthesis thin space left parenthesis 6 right parenthesis equals space 0 therefore space space space space 8 plus 4 minus straight lambda squared plus 72 space equals 0 space space space rightwards double arrow space space space space straight lambda squared space equals space 84 space space space space rightwards double arrow space space space straight lambda space equals space plus-or-minus square root of 84 space equals plus-or-minus square root of 4 cross times 21 end root therefore space space space space space space space straight lambda space equals plus-or-minus 2 square root of 21

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