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 $sin(\overrightarrow{p}\wedge \overrightarrow{q})$ is

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

Answer 1

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

$(3\overrightarrow{p}+q)\perp (5\overrightarrow{p}-3\overrightarrow{q})$

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

$15|\overrightarrow{p}{|}^2-3|\overrightarrow{q}{|}^2-4\overrightarrow{p}.\overrightarrow{q}=\mathrm{0....}\left(1\right)$

and $(2\overrightarrow{p}+\overrightarrow{q})\perp (4\overrightarrow{p}-2\overrightarrow{q})$

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

$8|\overrightarrow{p}{|}^2-2|\overrightarrow{q}{|}^2=0$

from $\left(1\right)$ and $\left(2\right)$

$15|\overrightarrow{p}{|}^2-3\times 4||{\overrightarrow{p}}^2-4|\overrightarrow{p}||q|\cos \theta =0$

$3|\overrightarrow{p}{|}^2-4|\overrightarrow{p}|2|\overrightarrow{p}|\cos \theta =0$

$\frac{3}{8}=\cos \theta$

$\sin \theta =\frac{\sqrt{5}}{8}$