If $a, b$ and $c$ are position vectors of the vertices of $\triangle ABC$,then $\frac{(a-c) \times (b-a)}{(b-a) \cdot (c-a)} = $

If $u = 2i + 2j - k$ and $v = 6i - 3j + 2k$,then find the unit vector perpendicular to both $u$ and $v$.

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 $\vec a = 2\hat i + \hat j - 2\hat k$ and $\vec b = \hat i + \hat j$. Let $\vec c$ be a vector such that $|\vec c - \vec a| = 3$,$|(\vec a \times \vec b) \times \vec c| = 3$,and the angle between $\vec c$ and $\vec a \times \vec b$ is $30^\circ$. Then $\vec a \cdot \vec c$ is equal to:

If $\vec{a}+\vec{b}+\vec{c}=0,$ show that $\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a} .$ Interpret the result geometrically.

If $a$ and $b$ are two non-zero perpendicular vectors,then a vector $y$ satisfying the equations $a \cdot y = c$ (where $c$ is a scalar) and $a \times y = b$ is