Q51 / 150 · MHT CET 2023 (12 May Shift 1) · Mathematics

If $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are three vectors, $|\overline{\mathrm{a}}|=2,|\overline{\mathrm{b}}|=4,|\overline{\mathrm{c}}|=1$, $|\overline{\mathrm{b}} \times \overline{\mathrm{c}}|=\sqrt{15}$ and $\overline{\mathrm{b}}=2 \overline{\mathrm{c}}+\lambda \overline{\mathrm{a}}$, then the value of $\lambda$, is

  • A
    $2$
  • B
    $2 \sqrt{2}$
  • C
    $1$
  • Correct answer
    D
    $4$

Solution

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

Hence, the correct answer is option: 'D'.

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