Let $\mathbf{a}=\mathbf{i}+2 \mathbf{j}+\mathbf{k}, \mathbf{b}=\mathbf{i}-\mathbf{j}+\mathbf{k},…
Let $\mathbf{a}=\mathbf{i}+2 \mathbf{j}+\mathbf{k}, \mathbf{b}=\mathbf{i}-\mathbf{j}+\mathbf{k}, \mathbf{c}=\mathbf{i}+\mathbf{j}-\mathbf{k}$. A vector in the plane of $\mathbf{a}$ and $\mathbf{b}$ has projection $\frac{1}{\sqrt{3}}$ on c. Then, one such vector is
$4 \mathbf{i}+\mathbf{j}-4 \mathbf{k}$
$3 \mathbf{i}+\mathbf{j}-3 \mathbf{k}$
$4 \mathbf{i}-\mathbf{j}+4 \mathbf{k}$
$2 \mathbf{i}+\mathbf{j}+2 \mathbf{k}$
Solution
Since, vectors $\mathbf{a}$ and $\mathbf{b}$ are in a same plane.
$\begin{aligned} \therefore \mathbf{r} & =\mathbf{a}+t \mathbf{b} \\ & =(\mathbf{i}+2 \mathbf{j}+\mathbf{k})+t(\mathbf{i}-\mathbf{j}+\mathbf{k}) \\ & =(1+t) \mathbf{i}+(2-t) \mathbf{j}+(1+t) \mathbf{k}\end{aligned}$
$\therefore$ Projection of $\mathbf{r}$ on $\mathbf{c}=\frac{\mathbf{r} \cdot \mathbf{c}}{|\mathbf{c}|}$
$\frac{1}{\sqrt{3}}=\frac{\left[\begin{array}{c}\{(1+t) \mathbf{i}+(2-t) \mathbf{j}+(1+t) \mathbf{k}\} \\ \cdot(\mathbf{i}+\mathbf{j}-\mathbf{k})\end{array}\right]}{\sqrt{1^2+1^2+(-1)^2}}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{1+t+2-t-(1+t)}{\sqrt{3}}$
$\Rightarrow \quad 1=2-t$
$\Rightarrow \quad t=1$
On putting $t=1$ in Eq. (i), we get $\mathbf{r}=2 \mathbf{i}+\mathbf{j}+2 \mathbf{k}$