Question

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

• A. 225
• B. 275
• C. 325
• D. 300

Answer 1

Answer: (D)

Since, $\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos 120^{\circ }$
$=1.2\cdot \left(-\frac{1}{2}\right)=-1$
$\because \{(\vec{a}+3\vec{b})\times (3\vec{a}-3\vec{b}){\}}^2$
$=\{3\vec{a}\times \vec{a}-\vec{a}\times \vec{b}+9\vec{b}\times \vec{a}-3\vec{b}\times \vec{b}{\}}^2$
$=[0-\vec{a}\times \vec{b}-9\vec{a}\times \vec{b}-0{]}^2$
$=100(\vec{a}\times \vec{b}{)}^2$
$=100\left\{a^2{b}^2-(\vec{a}\times \vec{b}{)}^2\right\}$
$=100\{4-1\}=300$