If $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are three vectors such that…

If $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are three vectors such that $\overline{\mathrm{a}} \neq \overline{0}$ and $\overline{\mathrm{a}} \times \overline{\mathrm{b}}=2 \overline{\mathrm{a}} \times \overline{\mathrm{c}},|\overline{\mathrm{a}}|=|\overline{\mathrm{c}}|=1,|\overline{\mathrm{~b}}|=4$ and $|\overline{\mathrm{b}} \times \overline{\mathrm{c}}|=\sqrt{15}$. If $\bar{b}-2 \bar{c}=\lambda \bar{a}$, then $\lambda$ is
  1. 1
  2. -4
  3. 3
  4. -2

Solution

If angle between $\overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ is $\alpha$ and $\begin{aligned} & |\overline{\mathrm{b}} \times \overline{\mathrm{c}}|=\sqrt{15} \\ & \Rightarrow|\overline{\mathrm{~b}}||\overline{\mathrm{c}}| \sin \alpha=\sqrt{15} \\ & \Rightarrow(4)(1) \sin \alpha=\sqrt{15} \\ & \Rightarrow \sin \alpha=\frac{\sqrt{15}}{4} \\ & \Rightarrow \cos \alpha=\frac{1}{4} \end{aligned}$ $\begin{aligned} & \text { Now, } \overline{\mathrm{b}}-2 \overline{\mathrm{c}}=\lambda \overline{\mathrm{a}} \\ & \Rightarrow|\overline{\mathrm{b}}-2 \overline{\mathrm{c}}|^2=\lambda^2|\overline{\mathrm{a}}|^2 \\ & \Rightarrow|\overline{\mathrm{~b}}|^2+4|\overline{\mathrm{c}}|^2-4 \overline{\mathrm{~b}} \cdot \overline{\mathrm{c}}=\lambda^2|\overline{\mathrm{a}}|^2 \\ & \Rightarrow 16+4-4(|\overline{\mathrm{~b}}| \overline{\mathrm{a}} \mid \cos \alpha)=\lambda^2 \\ & \Rightarrow 20-4\left(4 \times 1 \times \frac{1}{4}\right)=\lambda^2 \\ & \Rightarrow 16=\lambda^2 \\ & \Rightarrow \lambda= \pm 4\end{aligned}$

Asked in: MHT CET 2024 (16 May Shift 2)

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