Question

If the vectors $3\overline{p}+\overline{q};5\overline{p}-3\overline{q}$ and $2\overline{p}+\overline{q};4\overline{p}-2\overline{q}$ are pairs of mutually perpendicular vectors then $sin\left(\theta \right)$ is ($\theta$ is the angle between $\overline{p}$ and $\overline{q}$)

• A. $\sqrt{55}/4$
• B. $\sqrt{55}/8$
• C. $3/16$
• D. $\sqrt{247}/16$

Answer 1

The correct option is B $\sqrt{55}/8$

$(2\overrightarrow{p}+\overrightarrow{q}).(4\overrightarrow{p}-2\overrightarrow{q})=0$

$\Rightarrow 8p^2-2q^2=0$

$\Rightarrow q^2=4p^2$

$|q|=2|p|$

$(3\overrightarrow{p}+\overrightarrow{q}).(5\overrightarrow{p}-3\overrightarrow{q})=0$

$\Rightarrow 15p^2-3q^2-4\overrightarrow{p}.\overrightarrow{q}=0$

$4\overrightarrow{p}.\overrightarrow{q}=15p^2-3q^2$

$\Rightarrow \overrightarrow{p}.\overrightarrow{q}=\frac{3p^2}{4}$

Also, $|\overrightarrow{p}||\overrightarrow{q}|\cos \theta =\overrightarrow{p}.\overrightarrow{q}$

$\Rightarrow \cos \theta =\frac{\overrightarrow{p}.\overrightarrow{q}}{|\overrightarrow{p}||\overrightarrow{q}|}$

$\Rightarrow \cos \theta =\frac{3p^2}{4\times p\times 2p}=\frac{3}{8}$

$\therefore \sin \theta =\sqrt{1-{\cos }^2\theta }=\frac{\sqrt{55}}{8}$