Let $\bar{a}=\hat{i}+\hat{j}+\widehat{k}, \bar{b}=\hat{i}-\hat{j}+\widehat{k}$ and…
Let $\bar{a}=\hat{i}+\hat{j}+\widehat{k}, \bar{b}=\hat{i}-\hat{j}+\widehat{k}$ and $\bar{c}=\hat{i}-\hat{j}-\widehat{k}$ be three vectors. A vector $\overline{\mathrm{V}}$ in the plane of $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$, whose projection on $\overline{\mathrm{c}}$ is $\frac{1}{\sqrt{3}}$ is given by
$\hat{i}+3 \hat{j}-3 \widehat{k}$
$3 \hat{i}-\hat{j}+3 \hat{k}$
$\hat{i}-3 \hat{j}+3 \hat{k}$
$-3 \hat{i}-3 \widehat{j}-\widehat{k}$
Solution
Since, $v$ is the coplanar to $a$ and $b$
$\begin{aligned}
& \therefore v=a+t b^{-1} \\
& =(i+j+k)+t(i-j+k) \\
& \Rightarrow r=(1+t) i+(1-t) j+(1+t) k \ldots . . \text { (i) } \\
& \Rightarrow r=(1+t) l+(1-t) j+(1+t) k \text { (given) } \\
& \Rightarrow \frac{v \cdot c}{|c|}=\frac{1}{\sqrt{3}} \\
& \Rightarrow \frac{|(1+t) 1-1(1-t)-1(1+t)|}{\sqrt{3}}=\frac{1}{\sqrt{3}} \\
& \Rightarrow 1+t-1+t-1-t=1 \\
& \Rightarrow t=2
\end{aligned}$
On putting the value of $t$ in Eq. (i), we get
$\begin{aligned}
& r=3 i+(-1) j+(3) k \\
& \Rightarrow v=3 i-j+3 k
\end{aligned}$