#### If vectors a , b and c are non -coplanar unit vectors such that a cross (b cross c) = vec b+ vec c / square root 2 then the angle between vec a and vec b

Solution:   Since

$\vec{a}\times (\vec{b}\times \vec{c})=\frac{\vec{b}+\vec{c}}{\sqrt{2}}$

Therefore ,      $(\vec{a}\cdot \vec{c})\vec{b}-(\vec{a}\cdot \vec{b})\vec{c}=\frac{1}{\sqrt{2}}\vec{b}+\frac{1}{\sqrt{2}}\vec{c}$

Since    $\vec{b}$   and   $\vec{c}$  are non coplanar

$\\ \\ \Rightarrow \hspace{1cm}\vec{a}\cdot \vec{c}=\frac{1}{\sqrt{2}}\hspace{0.5cm}and \hspace{0.5cm}\vec{a}\cdot \vec{b}=\frac{1}{\sqrt{2}}\\ \\ \Rightarrow \hspace{0.5cm}\cos \theta =-\frac{1}{\sqrt{2}}\Rightarrow \theta=\frac{3\pi}{4}$