Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{x}, \vec{a} . \vec{x}=1, \vec{b} . \vec{x}=\frac{3}{2},|\vec{x}|=2$ Then angle between $\vec{c}$ and $\vec{x}$ is :
Answer & explanation
Correct answer: option 2
$\vec{a}+\vec{b}+\vec{c}=\vec{x}$
Taking dot with $\vec{x}$ on both sides, we get
$\vec{x} . \vec{a}+\vec{x} . \vec{b}+\vec{x} . \vec{c}+\vec{x} . \vec{x}=|\vec{x}|^2=4$
$\Rightarrow 1+\frac{3}{2}+\vec{x} . \vec{c}=4$
$\Rightarrow \vec{x} . \vec{c}=\frac{3}{2}$
If '$\theta$' be the angle between $\vec{c}$ and $\vec{x}$ then
$|\vec{x}||\vec{c}| \cos \theta=\frac{3}{2}$
$\Rightarrow \cos \theta=\frac{3}{4}$
$\Rightarrow \theta=\cos ^{-1}\left(\frac{3}{4}\right)$
Hence (2) is correct answer.