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Linear Combination of Vectors

If the vector...

Question

If the vector $6^i - 3^j - 6^k$ is decomposed into vectors parallel and perpendicular to the vector $^i +^j +^k$ then the vectors are :

A

- (^i +^j +^k) +^7 i -^2 j -^5 k

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B

- 2 (^i +^j +^k) +^8 i -^j -^4 k

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C

(^i +^j +^k) +^4 i -^5 j -^8 k

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D

n o n e

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Solution

The correct option is D $- (^i +^j +^k) +^7 i -^2 j -^5 k$
Given vector : $6 ˆ i - 3 ˆ j - 6 ˆ k$
Let $6 ˆ i - 3 ˆ j - 6 ˆ k$ = $c + d$ , where c is parallel to $\widehat i + \widehat j + \widehat k$ and d is perpendicular to $\widehat i + \widehat j + \widehat k$
hence $c = m\left( {\widehat i + \widehat j + \widehat k} \right)$ where m is a scalar
$d = a_{1} ˆ i + a_{2} ˆ j + a_{3} ˆ k$ where
$\begin{matrix} d \cdot (^i +^j +^k) = 0 s o a_{1} + a_{2} + a_{3} = 0 N o w 6^i - 3^j - 6^k = m (^i +^j +^k) + (a_{1}^i + a_{2}^j + a_{3}^k) s o e q u a t i o n c o e f f i c i e n t s, 6 = m + a_{1} 3 = m + a_{2} 6 = m + a_{3} \end{matrix}$
But $a_{1} + a_{2} + a_{3} = 0$ so $m = 1$
then $\begin{matrix} a_{1} = 7; a_{2} = - 2; a_{3} = - 5 6 ˆ i - 3 ˆ j - 6 ˆ k = c + d = m (ˆ i + ˆ j + ˆ k) + (a_{1} ˆ i + a_{2} ˆ j + a_{3} ˆ k) = - 1 (ˆ i + ˆ j + ˆ k) + (7 ˆ i - 2 ˆ j - 5 ˆ k) \end{matrix}$

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