Question

Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ be three vectors such that $|\overrightarrow{a}|=|\overrightarrow{c}|=1;|\overrightarrow{b}|=4$ and $|\overrightarrow{b}\times \overrightarrow{c}|=\sqrt{15}$. If $\overrightarrow{b}-2\overrightarrow{c}=\lambda \overrightarrow{a}$ then a value of $\lambda$ is

• A. $1$
• B. $-1$
• C. $2$
• D. $-4$

Answer 1

The correct option is D

$-4$

$(\overrightarrow{b}-2\overrightarrow{c})=\lambda \overrightarrow{a}\Rightarrow \frac{1}{\lambda }|\overrightarrow{b}-2\overrightarrow{c}|=1$

$\Rightarrow |\overrightarrow{b}-2\overrightarrow{c}{|}^2={\lambda }^2\Rightarrow |\overrightarrow{b}-2\overrightarrow{c}|.|\overrightarrow{b}-2\overrightarrow{c}|={\lambda }^2$

$\Rightarrow |\overrightarrow{b}{|}^2-4|\overrightarrow{b}||\overrightarrow{c}|.\cos \theta +4|\overrightarrow{c}{|}^2={\lambda }^2$

$\Rightarrow 16-4|\overrightarrow{b}||\overrightarrow{c}|.\cos \theta +4={\lambda }^2$

$\because |\overrightarrow{b}\times \overrightarrow{c}|=\sqrt{15}$

$\therefore |\overrightarrow{b}||\overrightarrow{c}|\sin \theta =\sqrt{15}$

$\Rightarrow \sin \theta =\frac{\sqrt{15}}{4}$

$\Rightarrow \cos \theta =\frac{1}{4}$

Thus, equation becomes $16-4\times \frac{1}{4}\times 4\times 1+4={\lambda }^2$

$\Rightarrow {\lambda }^2=16\Rightarrow \lambda =\pm 4$