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 :
Correct Option: , 2
$\overrightarrow{\mathrm{a}}=(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{c}}$
$=1.2 \cos \theta \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}$
$\Rightarrow \overrightarrow{\mathrm{a}}=2 \cos \theta \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}$
$|\overrightarrow{\mathrm{a}}|^{2}=(2 \cos \theta)^{2}+2^{2}-2.2 \cos \theta \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}$
$\Rightarrow 2=4 \cos ^{2} \theta+4-4 \cos \theta \cdot 2 \cos \theta$
$\Rightarrow-2=-4 \cos ^{2} \theta$
$\Rightarrow \cos ^{2} \theta=\frac{1}{2}$
$\Rightarrow \sec ^{2} \theta=2$
$\Rightarrow \tan ^{2} \theta=1$
$\Rightarrow \theta=\frac{\pi}{4}$
$1+\tan \theta=2$
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