Question

If the vector $6\hat{i}-3\hat{j}-6\hat{k}$ is decomposed into vectors parallel and perpendicular to the vector $\hat{i}+\hat{j}+\hat{k}$, then the vectors are

• A. $-(\hat{i}+\hat{j}+\hat{k})$ & $7\hat{i}-2\hat{j}-5\hat{k}$
• B. $-2(\hat{i}+\hat{j}+\hat{k})$ & $8\hat{i}-\hat{j}-4\hat{k}$
• C. $+2(\hat{i}+\hat{j}+\hat{k})$ & $8\hat{i}-\hat{j}-4\hat{k}$
• D. $none$

Answer 1

Answer: (A)

$-(\hat{i}+\hat{j}+\hat{k})$ & $7\hat{i}-2\hat{j}-5\hat{k}$

$\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$

And, $\overrightarrow{n}\perp \overrightarrow{a}$

Let,

$\overrightarrow{n}=x\hat{i}+y\hat{j}+z\hat{k}\overrightarrow{a}\cdot \overrightarrow{n}=0\Rightarrow x+y+z=0$

Also,

$\overrightarrow{a}+\overrightarrow{n}=6\hat{i}-3\hat{j}-6\hat{k}\Rightarrow x-1=6\Rightarrow x=7\Rightarrow y-1=-3\Rightarrow y=-2\Rightarrow z-1=-6\Rightarrow z=-5\Rightarrow \overrightarrow{n}=7\hat{i}-2\hat{j}-5\hat{k}$