Question

Let $\overrightarrow{a}=i+2\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{c}=i-\hat{j}-\hat{k}$. A vector in the plane of $\overrightarrow{a}$ and $\overrightarrow{b}$ whose projection on $\overrightarrow{c}$ is $\frac{1}{\sqrt{3}}$ is

• A. $4\hat{i}-\hat{j}+4\hat{k}$
• B. $3i+\hat{j}-3\hat{k}$
• C. $2i+\hat{j}-2\hat{k}$
• D. $4\hat{i}+\hat{j}-4\hat{k}$

Answer 1

Answer: (A)

$4\hat{i}-\hat{j}+4\hat{k}$

Vector lying in the plane of $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by, $\overrightarrow{r}={\lambda }_1\overrightarrow{a}+{\lambda }_2\overrightarrow{b}$
and its projection on $\overrightarrow{c}$ is $\frac{1}{\sqrt{3}}$
$\Rightarrow \left[\right({\lambda }_1+{\lambda }_2)\hat{i}-(2{\lambda }_1-{\lambda }_2)\hat{j}+({\lambda }_1+{\lambda }_2\left)\hat{k}\right]\cdot \frac{[\hat{i}-\hat{j}-\hat{k}]}{\sqrt{3}}=\frac{1}{\sqrt{3}}$
$\Rightarrow 2{\lambda }_1-{\lambda }_2=-1\Rightarrow \overrightarrow{r}=(3{\lambda }_1+1)\hat{i}-\hat{j}+(3{\lambda }_1+1)\hat{k}$
Hence, the required vector is, $4\hat{i}-\hat{j}+4\hat{k}$. corresponding to $\lambda =1$.