Question

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are such that $\left|\overrightarrow{a}\right|=1$ ,$\left|\overrightarrow{b}\right|,$ $\overrightarrow{a}.\overrightarrow{b}=2$ and $\overrightarrow{c}=2\overrightarrow{a}\times \overrightarrow{b}-3\overrightarrow{b}$, then the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is

• A. $\frac{\pi }{6}$
• B. $\frac{5\pi }{6}$
• C. $\frac{\pi }{3}$
• D. $\frac{2\pi }{3}$

Answer 1

Answer: (B)

$\frac{5\pi }{6}$

$\overrightarrow{a}\times \overrightarrow{b}$ is perpendicular to $\overrightarrow{b}$
${\left|\overrightarrow{c}\right|}^2={|2\overrightarrow{a}\times \overrightarrow{b}|}^2+{\left|3\overrightarrow{b}\right|}^2$
$=4{|\overrightarrow{a}\times \overrightarrow{b}|}^2+9{\left|\overrightarrow{b}\right|}^2$
$=4\times 12+9\times 4^2$
$=48+144=192$
$\therefore \left|\overrightarrow{c}\right|=8\sqrt{3}$
$\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{b}.(2\overrightarrow{a}\times \overrightarrow{b}-3\overrightarrow{b})$
$=-3\overrightarrow{b}.\overrightarrow{b}\therefore \overrightarrow{b}\times \overrightarrow{b}=0$
$=-3\times 16$
$=-48$
$\cos \theta =\frac{\overrightarrow{b}.\overrightarrow{c}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|}=\frac{-48}{4\times 8\sqrt{3}}=\frac{-\sqrt{3}}{2}$
$\cos \theta =\frac{\sqrt{3}}{2}\Rightarrow \theta =\frac{pi}{6}$
$\cos \theta$ is negative in second quadrants
$\theta =\pi -\frac{\pi }{6}$
$\therefore \theta =\frac{5\pi }{6}$