If two vectors $\vec{a}$ and $\vec{b}$ which are perpendicular to each other are such that $|\vec{a}|=8$ and $|\vec{b}|=3$, then $|\vec{a}-2\vec{b}|=$

Three non-coplanar vectors $\bar{a}, \bar{b}, \bar{c}$ are the coterminous edges of a parallelepiped. If $\bar{a}$ and $\bar{b}$ determine the base of the parallelepiped, then its height is:

If $a$ and $b$ are unit vectors such that $a+b$ is also a unit vector,then the angle between $a$ and $b$ is . . . . . . (in $^{\circ}$)

The projection of the vector $\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$ on the vector $\vec{b} = 4\hat{i} - 4\hat{j} + 7\hat{k}$ is:

Find the projection of the vector $\hat{i} - 2\hat{j} + \hat{k}$ on the vector $4\hat{i} - 4\hat{j} + 7\hat{k}$.

If $\vec{a}$ is a unit vector and $(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a}) = 8$,then $|\vec{x}| = $ . . . . . . .