Question

Let $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ be three vectors such that $\overrightarrow{a}=\overrightarrow{b}\times (\overrightarrow{b}\times \overrightarrow{c})$. If magnitudes of the vectors $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ are $\sqrt{2},1$ and $2$ respectively and the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is $\theta (0<\theta <\frac{\pi }{2})$, then the value of $1+\tan \theta$ is equal to

• A. $2$
• B. $\frac{\sqrt{3}+1}{\sqrt{3}}$
• C. $1$
• D. $\sqrt{3}+1$

Answer 1

The correct option is A $2$

$\overrightarrow{a}=\overrightarrow{b}\times (\overrightarrow{b}\times \overrightarrow{c})=(\overrightarrow{b}\cdot \overrightarrow{c})\overrightarrow{b}-{\left|\overrightarrow{b}\right|}^2\overrightarrow{c}$
$=(\overrightarrow{b}\cdot \overrightarrow{c})\overrightarrow{b}-\overrightarrow{c}(\because \left|\overrightarrow{b}\right|=1){\left|\overrightarrow{a}\right|}^2={(\overrightarrow{b}\cdot \overrightarrow{c})}^2{\left|\overrightarrow{b}\right|}^2+{\left|\overrightarrow{c}\right|}^2-2(\overrightarrow{b}\cdot \overrightarrow{c})(\overrightarrow{b}\cdot \overrightarrow{c})\Rightarrow 2={\left|\overrightarrow{c}\right|}^2-{(\overrightarrow{b}\cdot \overrightarrow{c})}^2\Rightarrow 2=4-(2\cos \theta {)}^2\Rightarrow (2\cos \theta {)}^2=2\Rightarrow \cos \theta =\frac{1}{\sqrt{2}}\Rightarrow \tan \theta =1$