Question

Let $\overrightarrow{a}=\overrightarrow{i}+\overrightarrow{j}+\hat{k}, \overrightarrow{b}=\hat{i}-\hat{j}+\hat{k} $ and $\overrightarrow{c}=\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k} $ be three vectors. A vector $\hat{v}$ in the plane of $\overrightarrow{a}$ and $ \overrightarrow{b},$ whose projection on $\overrightarrow{c} $ is $\frac{1}{\sqrt{3}},$ is given by

• A. $\widehat{i}-3\widehat{j}+3 \widehat{k}$
• B. $-3\widehat{i}-3\widehat{j} -\widehat{k}$
• C. $3\widehat{i}-\widehat{j}+ 3\widehat{k}$
• D. $\widehat{i}+3\widehat{j}- 3\widehat{k}$

Answer 1

The Correct Option is C

Let $\overrightarrow{v}=\overrightarrow{a}+\lambda\overrightarrow{b}$
$\overrightarrow{v}=(1+\lambda)\widehat{i}+(1+\lambda)\widehat{j}(1+\lambda)\widehat{k}$
Projection of $\overrightarrow{v} on \overrightarrow{c}=\frac{1}{\sqrt{3}}$
$\Rightarrow \hspace25mm \frac{\overrightarrow{v}.\overrightarrow{c}}{|\overrightarrow{c} | }=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{(1+\lambda)-(1-\lambda)-(1+\lambda)}{\sqrt{3}}=\frac{1}{\sqrt{3}}$
$\Rightarrow \hspace15mm 1+\lambda-1+\lambda-1-\lambda=1$
$\Rightarrow \hspace25mm \lambda-1=1$
$\Rightarrow \hspace25mm \lambda =2$
$\therefore \hspace25mm \overrightarrow{v}=3\widehat{i}-\widehat{j}+3\widehat{k}$