Question

Vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are inclined at an angle $\theta =120{}^{\circ }$ .If $|\overrightarrow{a}|=1,|\overrightarrow{b}|=2,$ then ${[(\overrightarrow{a}+3\overrightarrow{b})\times (3\overrightarrow{a}+\overrightarrow{b})]}^2$ is equal to

• A. 190
• B. 275
• C. 300
• D. 320
• E. 192

Answer 1

Answer: (E)

Given, $|\overrightarrow{a}|=1,|\overrightarrow{b}|=2$
$\therefore$ ${[(\overrightarrow{a}+3\overrightarrow{b})\times (3\overrightarrow{a}+\overrightarrow{b})]}^2$
$={[0+\overrightarrow{a}\times \overrightarrow{b}+9\overrightarrow{b}\times \overrightarrow{a}+0]}^2$
$={[-8\overrightarrow{a}\times \overrightarrow{b}]}^2$
$=64[|\overrightarrow{a}{|}^2|\overrightarrow{b}{|}^2{\sin }^2\theta ]$
$=64[1\times 4\times {\sin }^{2}120{}^{\circ }]$
$=64\times 4\times \frac{3}{4}$
$=192$