Vectors $\bar{i}$ and $\bar{b}$ are such that $|\bar{a}|=1 ;|\bar{b}|=4$ and $\bar{a} \cdot \bar{b}=2$. If…

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
  1. $\frac{\pi}{6}$
  2. $\frac{5 \pi}{6}$
  3. $\frac{\pi}{3}$
  4. $\frac{2 \pi}{3}$

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 $\cos \theta=\frac{(\bar{b} \cdot \bar{c})}{(|\mathrm{b} \| \mathrm{c}|)}=\frac{-48}{(4 \times 8 \sqrt{3})}$ $\cos \theta=\frac{-\sqrt{3}}{2}$ $\theta=\cos ^4\left(\frac{-\sqrt{3}}{2}\right)$ $\theta=\frac{5 \pi}{6}$

Asked in: MHT CET 2023 (09 May Shift 2)

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