Question

If $\overrightarrow{p}$ and $\overrightarrow{q}$ are two vectors with $|\overrightarrow{p}|=2$, $|\overrightarrow{q}|=3$, and the angle between them is $\frac{\pi }{3}$. If $\overrightarrow{x}$ is a vector such that $\overrightarrow{p}\times \overrightarrow{x}+2\overrightarrow{q}-3\overrightarrow{x}=0$, then vector $\overrightarrow{x}$ will be

• A. $\frac{2}{13}(\overrightarrow{p}+3\overrightarrow{q}+\overrightarrow{p}\times \overrightarrow{q})$
• B. $\frac{39}{2}(\overrightarrow{p}+\overrightarrow{q}-3(\overrightarrow{p}\times \overrightarrow{q}\left)\right)$
• C. $\frac{2}{39}(\overrightarrow{p}+\overrightarrow{q}+3(\overrightarrow{p}\times \overrightarrow{q}\left)\right)$
• D. Cannot be determined

Answer 1

The correct option is A $\frac{2}{13}(\overrightarrow{p}+3\overrightarrow{q}+\overrightarrow{p}\times \overrightarrow{q})$

Given that $|\overline{p}|=2,|\overline{q}|=3,\theta =\frac{\pi }{3}\Rightarrow \overline{p}.\overline{q}=3$
Consider $\overline{x}=\lambda \overline{p}+\mu \overline{q}+t(\overline{p}\times \overline{q})$
$\overline{p}\times \overline{x}=\mu (\overline{p}\times \overline{q})+t\left[\right(\overline{p}.\overline{q})\overline{p}-(\overline{p}.\overline{p}\left)\overline{q}\right]$
$\overline{p}\times \overline{x}=\mu (\overline{p}\times \overline{q})+t[3\overline{p}-4\overline{q}]$
We have,
$\overline{p}\times \overline{x}+2\overline{q}-3\overline{x}=0$
$\Rightarrow \mu (\overline{p}\times \overline{q})+t[3\overline{p}-4\overline{q}]+2\overline{q}-3(\lambda \overline{p}-\mu \overline{q}+t(\overline{p}\times \overline{q}\left)\right)=0$
$\Rightarrow (3t-3\lambda )\overline{p}+(2-4t-3\mu )\overline{q}+(\mu -3t)(\overline{p}\times \overline{q})=0$
$p,q,p\times q$ are non coplanar. Hence,
$3t-3\lambda =0$,
$2-4t-3\mu =0$
$\mu -3t=0$
On solving we get,
$\lambda =\frac{2}{13}$, $t=\frac{2}{13}$ and $\mu =\frac{6}{13}$
Therefore, $\overline{x}=\frac{2}{13}[\overline{p}+3\overline{q}+(p\times \overline{q}\left)\right]$
Hence, option 'A' is correct.