If $\vec{e_1}=\hat i+\hat j+\hat k$ and $\vec{e_2}=\hat i+\hat j-\hat k$ and $\vec a$ and $\vec b$ are two vectors such that $\vec{e_1}=2\vec a+\vec b$ and $\vec{e_2}=\vec a+2\vec b$, then angle between $\vec a$ and $\vec b$ is:
Answer & explanation
Correct answer: option 2
$2\vec a+\vec b=\hat i+\hat j+\hat k$
$\vec a+2\vec b=\hat i+\hat j-\hat k$
Solve to get:
$⇒\vec a=\frac{\hat i}{3}+\frac{\hat j}{3}+\hat k$ and $\vec b=\frac{\hat i}{3}+\frac{\hat j}{3}-\hat k$
$\cos θ=\frac{\vec a.\vec b}{|\vec a||\vec b|}=\frac{(\frac{1}{3})(\frac{1}{3})+(\frac{1}{3})(\frac{1}{3})+(1)(-1)}{(\sqrt{(\frac{1}{3})^2+(\frac{1}{3})^2+1})(\sqrt{(\frac{1}{3})^2+(\frac{1}{3})^2+1})}$
$⇒cos=\frac{-7}{11}⇒θ=cos^{-1}(\frac{-7}{11})$