If $a, b, c$ are position vectors of vertices of a triangle $ABC$,then the unit vector perpendicular to its plane is:

Let $\vec{a}$ and $\vec{b}$ be two non-zero vectors perpendicular to each other and $|\vec{a}|=|\vec{b}|$. If $|\vec{a} \times \vec{b}|=|\vec{a}|$,then the angle between the vectors $(\vec{a}+\vec{b}+(\vec{a} \times \vec{b}))$ and $\vec{a}$ is equal to

Find $|a \times b|^2$,if $|a|=2, |b|=3$ and the angle between $a$ and $b$ is $\theta = \frac{\pi}{6}$.

If $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar unit vectors such that $\bar{a} \times(\bar{b} \times \bar{c})=\frac{(\bar{b}+\bar{c})}{\sqrt{2}}$,then the angle between $\bar{a}$ and $\bar{b}$ is

Let $\vec{a}, \vec{b},$ and $\vec{c}$ be three unit vectors,out of which vectors $\vec{b}$ and $\vec{c}$ are non-parallel. If $\alpha$ and $\beta$ are the angles which vector $\vec{a}$ makes with vectors $\vec{b}$ and $\vec{c}$ respectively and $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{1}{2} \vec{b},$ then $|\alpha - \beta|$ is equal to .............. $^o$

Let $\bar{a}=4 \bar{i}+5 \bar{j}-\bar{k}$,$\bar{b}=\bar{i}-4 \bar{j}+5 \bar{k}$,$\bar{c}=3 \bar{i}+\bar{j}-\bar{k}$ and let $\bar{\alpha}$ be a vector perpendicular to both $\bar{a}$ and $\bar{b}$ such that $\bar{\alpha} \cdot \bar{c}=63$. Then $\bar{\alpha}=$