Q75 / 150 · MHT CET 2023 (09 May Shift 2) · Mathematics
Vectors $\bar{i}$ and $\bar{b}$ are such that $|\bar{a}|=1 ;|\bar{b}|=4$ and $\bar{a} \cdot \bar{b}-2$. If $\bar{c}-2 \bar{a} \times \bar{b}-3 \bar{b}$, then the angle between $\bar{b}$ and $\bar{c}$ is
- A$\frac{\pi}{6}$
- Correct answerB$\frac{5 \pi}{6}$
- C$\frac{\pi}{3}$
- D$\frac{2 \pi}{3}$
Solution
$\begin{aligned}
& \text { Given: }|\overline{\mathrm{a}}|=1,|\overline{\mathrm{b}}|=4 \text { and } \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=2 \text {, } \\
& \overline{\mathrm{c}}=2 \overline{\mathrm{a}} \times \overline{\mathrm{b}}-3 \overline{\mathrm{b}}
\end{aligned}$
Now that, $|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|^2=|\overline{\mathrm{a}}|^2|\overline{\mathrm{b}}|^2-|\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}|^2$
$\therefore \quad|\bar{a} \times \bar{b}|^2=16-4=12$
Given that $\overline{\mathrm{c}}=2 \overline{\mathrm{a}} \times \overline{\mathrm{b}}-3 \overline{\mathrm{b}}$
$\begin{aligned}
& |\bar{c}|^2=(2 \bar{a} \times \bar{b}-3 \bar{b})^2 \\
& |\bar{c}|^2=4|\bar{a} \times \bar{b}|^2+9|\bar{b}|^2
\end{aligned}$
$\begin{aligned}
& |\vec{c}|^2=4(12)+9(16) \\
& |\vec{c}|^2=192 \\
& |\vec{c}|=8 \sqrt{3}
\end{aligned}$
Now, $\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}=\mathrm{b} \cdot(2 \overline{\mathrm{a}} \times \overline{\mathrm{b}}-3 \overline{\mathrm{b}})$
$\therefore \quad \overline{\mathrm{b}} \cdot \overline{\mathrm{c}}=-3|\overline{\mathrm{b}}|^2=-48$
Angle between $b$ and $c$ is given by
$\cos \theta=\frac{(\bar{b} \cdot \bar{c})}{(|\mathrm{b} \| \mathrm{c}|)}=\frac{-48}{(4 \times 8 \sqrt{3})}$
$\cos \theta=\frac{-\sqrt{3}}{2}$
$\theta=\cos ^4\left(\frac{-\sqrt{3}}{2}\right)$
$\theta=\frac{5 \pi}{6}$
Hence, the correct answer is option: 'B'.