Question

Let vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ make an angle $\theta =\frac{2\pi }{3}$ between them. If $\left|\overrightarrow{a}\right|=1$ and $\left|\overrightarrow{b}\right|=2,$ then ${|(\overrightarrow{a}+3\overrightarrow{b})\times (3\overrightarrow{a}-\overrightarrow{b})|}^2$ is

• A. $300$
• B. $75$
• C. $400$
• D. $350$

Answer 1

The correct option is A $300$

${|(\overrightarrow{a}+3\overrightarrow{b})\times (3\overrightarrow{a}-\overrightarrow{b})|}^2$
$={|3\overrightarrow{a}\times \overrightarrow{a}-\overrightarrow{a}\times \overrightarrow{b}+9\overrightarrow{b}\times \overrightarrow{a}-3\overrightarrow{b}\times \overrightarrow{b}|}^2$
$={|3\times \overrightarrow{0}-\overrightarrow{a}\times \overrightarrow{b}-9\overrightarrow{a}\times \overrightarrow{b}-3\times \overrightarrow{0}|}^2$
$={|-10\overrightarrow{a}\times \overrightarrow{b}|}^2$
$=100{\left[\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\sin \theta \right]}^2$
$=100{\left[2\sin \frac{2\pi }{3}\right]}^2$
$=100{\left[2\sin (\pi -\frac{\pi }{3})\right]}^2$
$=100{\left[2\sin \frac{\pi }{3}\right]}^2$
$=100{(2\times \frac{\sqrt{3}}{2})}^2$
$=300$