Vector Algebra Question 14
Question 14 - 2024 (31 Jan Shift 1)
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
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Answer (48)
Solution
$\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=(2 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{b}}-3|\mathrm{~b}|^{2}$
$|\mathrm{b}||\mathrm{c}| \cos \alpha=-3|\mathrm{~b}|^{2}$
$|\mathrm{c}| \cos \alpha=-12$, as $|\mathrm{b}|=4$
$\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=2$
$\cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}$
$|c|^{2}=|(2 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})-3 \overrightarrow{\mathrm{b}}|^{2}$
$=64 \times \frac{3}{4}+144=192$
$|c|^{2} \cos ^{2} \alpha=144$
$192 \cos ^{2} \alpha=144$
$192 \sin ^{2} \alpha=48$
