If $P(3, 4, 5)$,$Q(4, 6, 3)$,$R(-1, 2, 4)$,and $S(1, 0, 5)$,then the projection of the vector $\vec{RS}$ on the vector $\vec{PQ}$ is:

If $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$,$|\vec{b}| = 5$,and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$,then the area of the triangle formed by these two vectors as two sides is

$A$ vector $a$ makes equal acute angles with the coordinate axes. Then the projection of vector $b = 5\hat{i} + 7\hat{j} - \hat{k}$ on $a$ is

Given that $\vec{a}, \vec{b}, \vec{p}$,and $\vec{q}$ are four vectors such that $\vec{a} + \vec{b} = \mu \vec{p}$,$\vec{b} \cdot \vec{q} = 0$,and $|\vec{b}|^2 = 1$,then $|(\vec{a} \cdot \vec{q}) \vec{p} - (\vec{p} \cdot \vec{q}) \vec{a}|$ is equal to:

If $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}$,$\vec{b}=\hat{i}-\hat{j}+3 \hat{k}$,$\vec{x}=\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right) \vec{b}$,$\vec{y}=\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right) \vec{a}$ and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$,then $x^2+y^2=$

The angle between the pair of lines with direction ratios $(1, 1, 2)$ and $(\sqrt{3} - 1, -\sqrt{3} - 1, 4)$ is ......... $^o$.