Question

If $|\overline{a}|=1,|\overline{b}|=2$, $(\overline{a},\overline{b})=\frac{2\pi }{3}$ then ${\left\{\right(\overline{a}+3\overline{b})\times (3\overline{a}-\overline{b}\left)\right\}}^2=?$

• A. $425$
• B. $375$
• C. $325$
• D. $300$

Answer 1

Answer: (D)

$300$

Given $|\overrightarrow{a}|=1,|\overrightarrow{b}|=2$, angle between $\overrightarrow{a}\times \overrightarrow{b}=\frac{2\pi }{3}=120°$. If $\overrightarrow{a},\overrightarrow{b}$ are two vectors then,

cross product $\overrightarrow{a}\times \overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\sin \theta$ and

dot product $\overrightarrow{a}.\overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\cos \theta$

Now, $\left\{\right(\overrightarrow{a}+3\overrightarrow{b})\times (3\overrightarrow{a}-\overrightarrow{b}){\}}^2=(\overrightarrow{a}\times 3\overrightarrow{b}+3\overrightarrow{b}\times 3\overrightarrow{a}-\overrightarrow{a}\times \overrightarrow{b}-3\overrightarrow{b}\times \overrightarrow{b}{)}^2$

$\Rightarrow (9\overrightarrow{b}\times \overrightarrow{a}-\overrightarrow{a}\times \overrightarrow{b}{)}^2[\overrightarrow{a}\times \overrightarrow{b}=-\overrightarrow{b}\times \overrightarrow{a}]$

$\Rightarrow (9\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{a})$

$\Rightarrow (10\overrightarrow{b}\times \overrightarrow{a})=(10|\overrightarrow{b}||\overrightarrow{a}|\sin 120°{)}^2$

$\Rightarrow (10\times 2\times \frac{\sqrt{3}}{2}{)}^2=300$