Question

Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ be three vectors such that $\overrightarrow{a}\ne 0$ and $|\overrightarrow{a}\mid =\mid \overrightarrow{c}\mid =1,\mid b\mid =4$ and $|\overrightarrow{b}\times \overrightarrow{c}\mid =\sqrt{15}$ ​, if $\overrightarrow{b}-2\overrightarrow{c}=\lambda \overrightarrow{a}$, then $\lambda$ equals to

• A. $1$
• B. $-1$
• C. $2$
• D. $-4$

Answer 1

The correct option is D $-4$

Let the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is $\alpha$
Then, $|\overrightarrow{b}\times \overrightarrow{c}|=\sqrt{15}$
$|\overrightarrow{b}||\overrightarrow{c}\mid \sin \alpha =\sqrt{15}$
$\sin \alpha =\frac{\sqrt{15}}{4}$
$\therefore \cos \alpha =\frac{1}{4}$
$\overrightarrow{b}-2\overrightarrow{c}=\lambda \overrightarrow{a}$
On squaring both sides,
$|\overrightarrow{b}-2\overrightarrow{c}{|}^2={\lambda }^2|\overrightarrow{a}{|}^2$
$|\overrightarrow{b}{|}^2+4|\overrightarrow{c}{|}^2-4\overrightarrow{b}\cdot \overrightarrow{c}={\lambda }^2|\overrightarrow{a}{|}^2$
$16+4-4\left(|\overrightarrow{b}||\overrightarrow{c}|\cos \alpha \right)={\lambda }^2\cdot (1{)}^2$
$20-4\times 4\times 1\times \frac{1}{4}={\lambda }^2$
$\Rightarrow {\lambda }^2=16$
$\Rightarrow \lambda =4\text{or}-4$