The position vectors of the vertices of a triangle $ABC$ are $4i - 2j$,$i + 4j - 3k$,and $-i + 5j + k$ respectively. Then $\angle ABC = $

If $\overrightarrow{a}=-\hat{i}+\hat{j}+2 \hat{k}$, $\overrightarrow{b}=2 \hat{i}-\hat{j}-\hat{k}$ and $\overrightarrow{c}=-2 \hat{i}+\hat{j}+3 \hat{k}$, then the angle between $2 \overrightarrow{a}-\overrightarrow{c}$ and $\overrightarrow{a}+\overrightarrow{b}$ is

$ABCD$ is a parallelogram and $P$ is a point on the segment $AD$ dividing it internally in the ratio $3:1$. If the line $BP$ meets the diagonal $AC$ in $Q$, then $AQ:QC$ equals

The cosine of the angle between any two diagonals of a cube is

If $\vec{a}$ and $\vec{b}$ are perpendicular unit vectors and vector $\vec{c}$ is such that $\vec{c} = \vec{a} + \vec{b}$,then $(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) + (\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) + (\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b})$ is

Let $a, b$ and $c$ be vectors with magnitudes $3, 4$ and $5$ respectively and $a + b + c = 0$. Then the value of $a \cdot b + b \cdot c + c \cdot a$ is: