Question

If the vectors $3\overrightarrow{p}+\overrightarrow{q},5\overrightarrow{p}-3\overrightarrow{q}$ and $2\overrightarrow{p}+\overrightarrow{q},4\overrightarrow{p}-2\overrightarrow{q}$ are pairs of mutually perpendicular vectors, then the angle between vectors $\overrightarrow{p}$ and $\overrightarrow{q}$ is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{2}$
• C. ${\cos }^{-1}\left(\frac{1}{4}\right)$
• D. ${\cos }^{-1}\left(\frac{3}{8}\right)$

Answer 1

The correct option is D ${\cos }^{-1}\left(\frac{3}{8}\right)$

Given : $3\overrightarrow{p}+\overrightarrow{q},5\overrightarrow{p}-3\overrightarrow{q}$ and $2\overrightarrow{p}+\overrightarrow{q},4\overrightarrow{p}-2\overrightarrow{q}$ are pairs of mutually perpendicular vectors.
Let $\theta$ be the angle between $\overrightarrow{p}$ and $\overrightarrow{q}$

$\Rightarrow (3\overrightarrow{p}+\overrightarrow{q})\cdot (5\overrightarrow{p}-3\overrightarrow{q})=0$
$\Rightarrow 15|\overrightarrow{p}{|}^2-3|\overrightarrow{q}{|}^2-4\overrightarrow{p}\cdot \overrightarrow{q}=0\cdots \left(i\right)$
and
$\Rightarrow (2\overrightarrow{p}+\overrightarrow{q})\cdot (4\overrightarrow{p}-2\overrightarrow{q})=0$
$\Rightarrow 8|\overrightarrow{p}{|}^2-2|\overrightarrow{q}{|}^2=0$
$\Rightarrow |\overrightarrow{q}|=2|\overrightarrow{p}|$
Using this in $\left(i\right),$ we get
$15|\overrightarrow{p}{|}^2-12|\overrightarrow{p}{|}^2-4|\overrightarrow{p}|\times 2|\overrightarrow{p}|\times \cos \theta =0$
$\Rightarrow \cos \theta =\frac{3}{8}$
$\Rightarrow \theta ={\cos }^{-1}\left(\frac{3}{8}\right)$