If $a$ and $b$ are mutually perpendicular vectors,then $(a + b)^2 = $

Vector $\vec{a} + 3\vec{b}$ is perpendicular to $7\vec{a} - 5\vec{b}$ and $\vec{a} - 5\vec{b}$ is perpendicular to $7\vec{a} + 3\vec{b}$. The angle between non-zero vectors $\vec{a}$ and $\vec{b}$ is:

$\overline{u}, \overline{v}, \overline{w}$ 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 $|\overline{u}-\overline{v}+\overline{w}|=$

Let $\vec{a}=\hat{i}-2\hat{j}+3\hat{k}$, $\vec{b}=2\hat{i}+\hat{j}-\hat{k}$, $\vec{c}=\lambda\hat{i}+\hat{j}+\hat{k}$ and $\vec{v}=\vec{a}\times\vec{b}$. If $\vec{v} \cdot \vec{c}=11$ and the length of the projection of $\vec{b}$ on $\vec{c}$ is $p$, then $9p^{2}$ is equal to:

Let the position vectors of the vertices $A, B$ and $C$ of a tetrahedron $ABCD$ be $\hat{i}+2\hat{j}+\hat{k}$,$\hat{i}+3\hat{j}-2\hat{k}$ and $2\hat{i}+\hat{j}-\hat{k}$ respectively. The altitude from the vertex $D$ to the opposite face $ABC$ meets the median line segment through $A$ of the triangle $ABC$ at the point $E$. If the length of $AD$ is $\frac{\sqrt{110}}{3}$ and the volume of the tetrahedron is $\frac{\sqrt{805}}{6\sqrt{2}}$,then the position vector of $E$ is

The angle between the vectors $\bar{a} = 6 \hat{i} + 2 \hat{j} - 8 \hat{k}$ and $\bar{b} = 4 \hat{i} - 4 \hat{j} + 2 \hat{k}$ is . . . . . . .