Q80 / 200 · Manipal (MET) 2011 · Mathematics

Let $\mathbf{a}, \mathbf{b}, \mathbf{c}$, be three vectors such that $\mathbf{c} \neq 0$ and $\mathbf{a} \cdot \mathbf{b} \quad=2 \mathbf{a} \cdot \mathbf{c},|\mathbf{a}|=|\mathbf{c}|=1,|\mathbf{b}|=4 \quad$ and $|b \times c|=\sqrt{15}$, if $b-2 c=\lambda \alpha$. then, $\lambda$ equals

  • A
    -1
  • B
    1
  • Correct answer
    C
    -4
  • D
    2

Solution

Let angle between $\mathbf{b}$ and $\mathbf{c}$ is $\alpha$, then $|\mathbf{b} \times \mathbf{c}|=\sqrt{15}$ (given) $\begin{aligned} & \Rightarrow \quad|b||c| \sin \alpha=\sqrt{15} \\ & \Rightarrow \quad \sin \alpha=\frac{\sqrt{15}}{4}\end{aligned}$ $\therefore \quad \cos \alpha=\frac{1}{4}$ Also, $\quad \mathbf{b}-2 \mathrm{c}=\lambda \mathbf{a}$ $\begin{array}{ll}\Rightarrow & (b-2 c)^2=\lambda^2(a)^2 \\ \Rightarrow & |b|^2+4|c|^2-4 b \cdot c=\lambda^2|a|^2 \\ \Rightarrow & 16+4-4\{|b||c| \cos \alpha\}=\lambda^2 \\ \Rightarrow & \lambda^2=16 \\ \Rightarrow & \lambda= \pm 4\end{array}$

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

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