Question

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are three vectors such that $\left|\overrightarrow{a}\right|=\left|\overrightarrow{c}\right|=1,\left|\overrightarrow{b}\right|=4,|\overrightarrow{b}\times \overrightarrow{c}|=2$ and $2\overrightarrow{b}=\overrightarrow{c}+\lambda \overrightarrow{a}$ then $\lambda$ is

• A. $\sqrt{65-8\sqrt{3}}$
• B. $\sqrt{17}$
• C. $\sqrt{3}$
• D. $\sqrt{\frac{17}{2}(2+\sqrt{3})}$

Answer 1

The correct option is A $\sqrt{65-8\sqrt{3}}$

$|\overrightarrow{b}\times \overrightarrow{c}|=\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|\sin \theta ,\theta$ being the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$.
$\Rightarrow 2=4\sin \theta$
$\Rightarrow \theta =\frac{\pi }{6}$

Now, we have $\lambda \overrightarrow{a}=2\overrightarrow{b}-\overrightarrow{c}$
$\Rightarrow \lambda |\overrightarrow{a}|=\sqrt{4|\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2-4\overrightarrow{b}\cdot \overrightarrow{c}}$
$\Rightarrow {\lambda }^2=64+1-8\sqrt{3}$
$\Rightarrow \lambda =\sqrt{65-8\sqrt{3}}$