Vectors $\vec a$ and $\vec b$ are inclined at an angle θ = 60°. If $|\vec a|=1,|\vec b|=2$, then $[(\vec a + 3\vec b) × (3\vec a -\vec b)]^2$ is equal to
Answer & explanation
Correct answer: option 4
We have,
$[(\vec a + 3\vec b) × (3\vec a -\vec b)]^2$
$=[10 (\vec b×\vec a)]^2$
$=100|\vec b×\vec a|^2$
$=100\left\{|\vec a|^2|\vec b|^2-(\vec a.\vec b)^2\right\}$
$=100(4-2\cos 60°)=300$ $[∵\vec a.\vec b=2\cos 60°=1$