Question

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}| = 1$, $|\vec{b}| = 4$ and $\vec{a} \cdot \vec{b} = 2$.If $\vec{c} = (2 \vec{a} \times \vec{b}) - 3 \vec{b}$ and the angle between $\vec{b}$ and $\vec{c}$ is $\alpha$, then $192 \sin^2 \alpha$ is equal to _____

Answer 1

Given:

$\vec{b} \cdot \vec{c} = \left( 2\vec{a} \times \vec{b} \right) \cdot \vec{b} - 3|\vec{b}|^2$

We know:

$|\vec{b}||\vec{c}| \cos \alpha = -12, \quad \text{as } |\vec{b}| = 4, \; \vec{a} \cdot \vec{b} = 2$

Calculate:

$\cos \alpha = \frac{1}{2}, \quad \alpha = \frac{\pi}{3}$

Now:

$|\vec{c}|^2 = |(2\vec{a} \times \vec{b}) - 3\vec{b}|^2 = 64 \times \frac{3}{4} + 144 = 192$

Therefore:

$192 \cos^2 \alpha = 144, \quad 192 \sin^2 \alpha = 48$