If $a \cdot b = 0$,then:

Let $u$ and $v$ be two non-zero vectors. If $|u+v|=|u-v|$,then:

The position vectors of the vertices of a $\Delta ABC$ are $4\hat{i}-2\hat{j}$,$\hat{i}+4\hat{j}-3\hat{k}$,and $-\hat{i}+5\hat{j}+\hat{k}$ respectively. Then,$\angle ABC$ is equal to:

If $4 \hat{i}+7 \hat{j}+8 \hat{k}$,$2 \hat{i}+3 \hat{j}+4 \hat{k}$,and $2 \hat{i}+5 \hat{j}+7 \hat{k}$ are respectively the position vectors of the vertices $A, B, C$ of $\triangle ABC$,then the position vector of the point where the bisector of angle $A$ meets $BC$ is

$A$ tetrahedron has vertices at $O(0, 0, 0)$,$A(1, 2, 1)$,$B(2, 1, 3)$,and $C(-1, 1, 2)$. The angle between the faces $OAB$ and $ABC$ is:

If the position vectors of the vertices of $\Delta ABC$ are $2\hat{i} + 4\hat{j} - \hat{k}$,$4\hat{i} + 5\hat{j} + \hat{k}$,and $3\hat{i} + 6\hat{j} - 3\hat{k}$,then which of the following angles is a right angle?