The projection of $\overline{AB}$ on $\overline{CD}$,where $A \equiv (2, -3, 0)$,$B \equiv (1, -4, -2)$,$C \equiv (4, 6, 8)$,and $D \equiv (7, 0, 10)$ is

If $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$,$\vec{b}=-\hat{i}+2\hat{j}+\hat{k}$,$\vec{c}=3\hat{i}+\hat{j}$ and $\vec{a}+\lambda\vec{b}$ is perpendicular to $\vec{c}$,then $\lambda=$

$\overline{u}, \overline{v}, \overline{w}$ are three vectors such that $|\overline{u}|=1, |\overline{v}|=2, |\overline{w}|=3$. If the projection of $\overline{v}$ along $\overline{u}$ is equal to the projection of $\overline{w}$ along $\overline{u}$ and $\overline{v}, \overline{w}$ are perpendicular to each other,then $|\overline{u}-\overline{v}+\overline{w}|=$

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:

If $|\vec{a}|=4$ and $|\vec{b}|=5$,then the values of $k$ for which $\vec{a}+k \vec{b}$ is perpendicular to $\vec{a}-k \vec{b}$ are

Two adjacent sides of a parallelogram $ABCD$ are given by $\vec{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\vec{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$. The side $\vec{AD}$ is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that $\vec{AD}$ becomes $\vec{AD'}$. If $\vec{AD'}$ makes a right angle with the side $\vec{AB}$, then the cosine of the angle $\alpha$ is given by