Solve this following
Question:

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

of $1+\tan \theta$ is equal to :

1. $\sqrt{3}+1$

2. 2

3. 1

4. $\frac{\sqrt{3}+1}{\sqrt{3}}$

Correct Option: , 2

Solution:

$\overrightarrow{\mathrm{a}}=(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{c}}$

$=1.2 \cos \theta \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}$

$\Rightarrow \overrightarrow{\mathrm{a}}=2 \cos \theta \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}$

$|\overrightarrow{\mathrm{a}}|^{2}=(2 \cos \theta)^{2}+2^{2}-2.2 \cos \theta \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}$

$\Rightarrow 2=4 \cos ^{2} \theta+4-4 \cos \theta \cdot 2 \cos \theta$

$\Rightarrow-2=-4 \cos ^{2} \theta$

$\Rightarrow \cos ^{2} \theta=\frac{1}{2}$

$\Rightarrow \sec ^{2} \theta=2$

$\Rightarrow \tan ^{2} \theta=1$

$\Rightarrow \theta=\frac{\pi}{4}$

$1+\tan \theta=2$