Q54 / 125 · BITSAT 2015 · Mathematics
Let $a, b$ and $c$ be three vectors satisfying $a \times b=$ $(a \times c),|a|=|c|=1,|b|=4$ and $|b \times c|=\sqrt{15}$. If $b-$
$2 \mathrm{c}=\lambda \mathrm{a},$ then $\lambda$ equals
- A1
- B-1
- C2
- Correct answerD-4
Solution
Let $\theta$ be the angle between $\mathbf{b}$ and $\mathbf{c}$. Given, $|\mathrm{b} \times \mathrm{c}|=\sqrt{15}$
$$
\begin{array}{l}
\Rightarrow|\mathrm{b} \| \mathrm{c}| \sin \theta=\sqrt{15} \\
\Rightarrow \sin \theta=\frac{\sqrt{15}}{|\mathrm{~b}| \mathrm{c} \mid} \\
\Rightarrow \sin \theta=\sqrt{\frac{15}{4 \times 1}}=\frac{\sqrt{15}}{4} \\
\quad \therefore \cos \theta=\sqrt{1-\frac{15}{16}}=\frac{1}{\sqrt{16}}=\frac{1}{4}
\end{array}
$$
Now given, $\mathrm{b}-2 \mathrm{c}=\lambda \mathrm{a} \Rightarrow\left|\mathrm{b}-2 \mathrm{c}^{2}\right|=\mid \lambda \mathrm{a}^{2}$
$$
\Rightarrow|\mathrm{b}|^{2}+4|\mathrm{c}|-4(\mathrm{~b} \cdot \mathrm{c})=\lambda^{2}|\mathrm{a}|^{2}
$$
$$
\begin{array}{l}
\Rightarrow 16+4-4|\mathrm{~b} \| \mathrm{c}| \cos \theta=\lambda^{2} \\
\Rightarrow 20-16 \cos \theta=\lambda^{2} \\
\Rightarrow 20-4=\lambda^{2} \Rightarrow \lambda^{2}=16\left(\therefore \cos \theta=\frac{1}{4}\right) \\
\Rightarrow \lambda=\pm 4
\end{array}
$$
Hence, the correct answer is option: 'D'.