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Vector Algebra

Question
CBSEENMA12034111
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Express the vector straight a with rightwards arrow on top space equals space 5 space straight i with hat on top space minus space 2 space straight j with hat on top space plus space 5 space straight k with hat on top as sum of two vectors such that one is parallel to the vector straight b with rightwards arrow on top space equals space 3 space straight i with hat on top space plus space straight k with hat on top and other is perpendicular to straight b with rightwards arrow on top.

Solution

                       straight a with rightwards arrow on top space equals space 5 straight i with hat on top space minus space 2 straight j with hat on top plus 5 straight k with hat on top comma space space space straight b with rightwards arrow on top space equals 3 straight i with hat on top space plus straight k with hat on top
Let straight a with rightwards arrow on top space equals space straight A with rightwards arrow on top space plus space straight B with rightwards arrow on top space where space straight A with rightwards arrow on top space is space parallel space to space straight b with rightwards arrow on top space and space straight B with rightwards arrow on top space is space perpendicular space to space straight b with rightwards arrow on top.
therefore space space space space straight A with rightwards arrow on top space equals space straight lambda left parenthesis 3 space straight i with hat on top space plus space straight k with hat on top right parenthesis space equals space 3 space straight lambda space straight i with hat on top space plus space straight lambda space straight k with hat on top
and straight B with rightwards arrow on top space equals space straight a with rightwards arrow on top space minus space straight A with rightwards arrow on top space equals space left parenthesis 5 straight i with hat on top plus straight j with hat on top space plus space 5 space straight k with hat on top right parenthesis space minus space left parenthesis 3 space straight lambda space straight i with hat on top space plus space straight lambda space straight k with hat on top right parenthesis
            equals space left parenthesis 5 space minus 3 space straight lambda right parenthesis space straight i with hat on top space plus space straight j with hat on top space plus space left parenthesis 5 minus straight lambda right parenthesis space straight k with hat on top
Since straight B with rightwards arrow on top is perpendicular to straight b with rightwards arrow on top
therefore space space space space straight B with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 0 rightwards double arrow space space space left parenthesis 5 space minus space 3 space straight lambda right parenthesis space left parenthesis 3 right parenthesis space plus space left parenthesis 1 right parenthesis thin space left parenthesis 0 right parenthesis space plus space left parenthesis 5 minus straight lambda right parenthesis space left parenthesis 1 right parenthesis space equals space 0 rightwards double arrow space space space space 15 minus 9 straight lambda plus 5 minus straight lambda space equals space 0 space space space space space space space space space space space space space space space rightwards double arrow space space 10 straight lambda space equals space 20 space space space space rightwards double arrow space space space straight lambda space equals space 2 therefore space space space space space space straight A with rightwards arrow on top space equals space 6 straight i with hat on top space plus space 2 straight k with hat on top comma space space space straight B with rightwards arrow on top space equals space minus straight i with hat on top plus space straight j with hat on top space plus space 3 straight k with hat on top


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