Question

$\overrightarrow{a}=-2\hat{i}+\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}+5\hat{j}and\overrightarrow{c}=4\hat{i}+4\hat{j}+2\hat{k}$ the projection of the vector $3\overrightarrow{a}-2\overrightarrow{b}$ on the vector $\overrightarrow{c}$ is

• A. 11
• B. -11
• C. 13
• D. none of these

Answer 1

The correct option is D none of these

Given vectors,

$\overrightarrow{a}=-2\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}$

$\overrightarrow{b}=\overrightarrow{i}+5\overrightarrow{j}$

$\overrightarrow{c}=4\overrightarrow{i}+4\overrightarrow{j}+2\overrightarrow{k}$

$\Rightarrow 3\overrightarrow{a}-2\overrightarrow{b}=3(-2\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k})-2(\overrightarrow{i}+5\overrightarrow{j})$

$=-6\overrightarrow{i}+3\overrightarrow{j}+3\overrightarrow{k}-2\overrightarrow{i}-10\overrightarrow{j}$

$=-8\overrightarrow{i}-7\overrightarrow{j}+3k$

$\Rightarrow$ Now unit vector along $\overrightarrow{c}$ will be given as

$\hat{c}=\frac{\overrightarrow{c}}{|\overrightarrow{c}|}=\frac{4\hat{i}+4\hat{j}+2\hat{k}}{\sqrt{4^2+4^2+2^2}}$

$\frac{4}{6}\hat{i}+\frac{4}{6}\hat{j}+\frac{2}{6}\hat{k}$

$=(3\overrightarrow{a}-2\overrightarrow{b})\widehat{.c}$

$(-8\hat{i}-7\hat{j}+3\hat{k}).(\frac{4}{6}\hat{i}+\frac{4}{6}\hat{j}+\frac{2}{6}\hat{k})$

$=-8\times \frac{4}{6}-7\times \frac{4}{6}+3\times \frac{2}{6}$

$=\frac{-32}{6}-\frac{28}{6}+\frac{6}{6}$

$=\frac{-16}{3}-\frac{14}{3}+1$

$=\frac{-16-14+3}{3}=\frac{-27}{3}=-9$

Ans is $\left(D\right)$ None of these