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Vector Algebra

Question
CBSEENMA12034112

Decompose the vector 6 straight i with hat on top space minus space 3 straight j with hat on top space minus space 6 straight k with hat on top into vectors which are parallel and perpendicular to the vector straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top.

Solution

Let straight a with rightwards arrow on top space equals space 6 straight i with hat on top space minus space 3 straight j with hat on top minus space 6 straight k with hat on top be decomposed into two vectors straight A with rightwards arrow on top space and space 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 space straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat 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 straight A with rightwards arrow on top space equals space straight lambda space left parenthesis straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top right parenthesis space equals space straight lambda space straight i with hat on top space plus space straight lambda space straight j with hat on top space plus space straight lambda space straight k with hat on top
and     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 space space space space space space space space space space space space space space space space rightwards double arrow space space space space 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
therefore space space space straight B with rightwards arrow on top space equals space left parenthesis 6 minus space straight lambda right parenthesis space straight i with hat on top space minus space left parenthesis straight lambda plus 3 right parenthesis space straight j with hat on top space minus space left parenthesis straight lambda minus 6 right parenthesis space straight k with hat on top
because space 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 b with rightwards arrow on top. space straight B with rightwards arrow on top space equals space 0
rightwards double arrow space space left parenthesis 6 minus straight lambda right parenthesis space left parenthesis 1 right parenthesis space minus space left parenthesis straight lambda plus 3 right parenthesis space left parenthesis 1 right parenthesis space minus space left parenthesis straight lambda plus 6 right parenthesis space left parenthesis 1 right parenthesis space equals space 0
therefore space space space space 6 space minus space straight lambda space minus straight lambda minus 3 minus straight lambda minus 6 space equals space 0 space space space space space space space rightwards double arrow space space space space space 3 space straight lambda space equals space minus 3 space space space space rightwards double arrow space space space straight lambda space equals space minus 1
therefore space space space required space vectors space are
space space space space space space space space space space space space space minus straight i with hat on top space minus space straight j with hat on top space minus space straight k with hat on top comma space space space space space 7 straight i with hat on top space minus space 2 space straight j with hat on top space minus space 5 space straight k with hat on top