JEE Main 2021 (Online) 27th July Evening ShiftQuestion 1
Let $\vec a,\vec b$ and $\vec c$ be three vectors such that $\vec a=\vec b\times(\vec b\times\vec c)$. If magnitudes of the vectors $\vec a,\vec b$ and $\vec c$ are $\sqrt2,1$ and $2$ respectively and the angle between $\vec b$ and $\vec c$ is $\theta$ $\left(0<\theta<\frac\pi2\right)$, then the value of $1+\tan\theta$ is equal to :
Vector Algebra
Solution
$\vec a=\vec b\times(\vec b\times\vec c)=\vec b(\vec b\cdot\vec c)-\vec c|\vec b|^2$. $|\vec a|^2=(\vec b\cdot\vec c)^2-2(\vec b\cdot\vec c)^2+|\vec c|^2=4-4\cos^2\theta=4\sin^2\theta$. $4\sin^2\theta=2\Rightarrow\sin^2\theta=\frac12$, $\theta=\frac\pi4$ (acute). $1+\tan\theta=2$. Option (B).