Mathematics - Vector Question with Solution | TestHub

Since vector $\overrightarrow{v}, \overrightarrow{a} \& \overrightarrow{b}$ are coplanar then,  $\overrightarrow{v}={\lambda }_1\overrightarrow{a}+{\lambda }_2\overrightarrow{b}$, where ${\lambda }_1,{\lambda }_2\in \mathrm{ℝ}$.

$\Rightarrow \overrightarrow{v}=\left({\lambda }_1+2{\lambda }_2\right)\hat{i}+\left({\lambda }_1-3{\lambda }_2\right)\hat{j}+\left(2{\lambda }_1+{\lambda }_2\right)\hat{k}$

$\because$ Projection of $\overrightarrow{v}$ on $\overrightarrow{c}$ is $\frac{2}{\sqrt{3}}$

$\therefore \frac{{\lambda }_1+2{\lambda }_2-{\lambda }_1+3{\lambda }_2+2{\lambda }_1+{\lambda }_2}{\sqrt{3}}=\frac{2}{\sqrt{3}}$

$\therefore {\lambda }_1+3{\lambda }_2=1 ...\left(1\right)$

and $\overrightarrow{v}·\hat{j}=7\Rightarrow {\lambda }_1-3{\lambda }_2=7 ...\left(2\right)$

$\Rightarrow \overrightarrow{v}=\left({\lambda }_1+2{\lambda }_2\right)\hat{i}+\left({\lambda }_1-3{\lambda }_2\right)\hat{j}+\left(2{\lambda }_1+{\lambda }_2\right)\hat{k}$

$\therefore {\lambda }_1+3{\lambda }_2=1 ...\left(1\right)$

$\Rightarrow {\lambda }_1-3{\lambda }_2=7 \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$, we get

${\lambda }_1=4,{\lambda }_2=-1$

$\therefore \overrightarrow{v}=2\hat{i}+7\hat{j}+7\hat{k}$

$\therefore \overrightarrow{v}·\left(\hat{i}+\hat{k}\right)=2+7=9$