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Vector Algebra

Question
CBSEENMA12034113

Express 2 straight i with hat on top space minus straight j with hat on top space plus space 3 straight k with hat on top as the sum of a vector parallel, and a vector parallel, and a vector perpendicular to 2 straight i with hat on top plus 4 straight j with hat on top minus 2 straight k with hat on top.

Solution

Let straight a with rightwards arrow on top space equals space 2 straight i with hat on top space minus space straight j with hat on top plus 3 straight k with hat on top be expressed as
            straight a with rightwards arrow on top space equals space straight A with rightwards arrow on top space plus straight B with rightwards arrow on top
where straight A with rightwards arrow on top is parallel to straight b with rightwards arrow on top space equals 2 straight i with hat on top plus 4 straight j with hat on top minus 2 straight k with hat on top and B is perpendicular to straight b with rightwards arrow on top.
therefore space space space straight A with rightwards arrow on top space equals space straight lambda left parenthesis 2 straight i with hat on top plus 4 straight j with hat on top minus 2 straight k with hat on top right parenthesis space equals space 2 straight lambda straight i with hat on top space plus space 4 straight lambda straight j with hat on top space minus 2 straight lambda straight k with hat on top
and space straight B with rightwards arrow on top space equals space straight a with rightwards arrow on top minus straight A with rightwards arrow on top space equals left parenthesis 2 minus 2 straight lambda right parenthesis straight i with hat on top space minus space left parenthesis 1 plus 4 straight lambda right parenthesis straight j with hat on top space plus space left parenthesis 3 plus 2 straight lambda right parenthesis straight k with hat on top
therefore space straight B with rightwards arrow on top space is space perpendicular space to space straight b with rightwards arrow on top
because space space straight B with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 0
therefore space space space space left parenthesis 2 minus 2 straight lambda right parenthesis space left parenthesis 2 right parenthesis space minus space left parenthesis 1 plus 4 straight lambda right parenthesis space left parenthesis 4 right parenthesis space plus space left parenthesis 3 plus 2 straight lambda right parenthesis space left parenthesis negative 2 right parenthesis space equals space 0
therefore space space space space 4 minus 4 straight lambda minus 4 minus 16 straight lambda minus 6 minus 4 straight lambda space equals 0
rightwards double arrow space space space 24 straight lambda space equals space minus 6 space space space space space space space space space space space space space space space rightwards double arrow space straight lambda space equals space minus 1 fourth
therefore space space space required space vectors space are
space space space space space space space space minus 1 half straight i with hat on top space minus space straight j with hat on top space plus space 1 half straight k with hat on top space and space 5 over 2 straight i with hat on top space plus space 5 over 2 straight k with hat on top
space space therefore space space space 2 space straight i with hat on top space minus space straight j with hat on top space plus space 3 space straight k with hat on top space equals space open parentheses negative 1 half straight i with hat on top space minus space straight j with hat on top space plus space 1 half straight k with hat on top close parentheses space plus space open parentheses 5 over 2 straight i with hat on top space plus space 5 over 2 straight k with hat on top close parentheses