If vectors $\vec{a}$ and $\vec{b}$ have the same magnitude, the angle between them is $60^\circ$, and their scalar product is $1/2$, then $|\vec{a}|$ is:

Let $\overrightarrow{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$ and $\overrightarrow{b} = 7\hat{i} + \hat{j} - 6\hat{k}$. If $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{r} \times \overrightarrow{b}$ and $\overrightarrow{r} \cdot (\hat{i} + 2\hat{j} + \hat{k}) = -3$,then $\overrightarrow{r} \cdot (2\hat{i} - 3\hat{j} + \hat{k})$ is equal to:

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

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 values of $a$,for which points $A, B, C$ with position vectors $2\hat{i}-\hat{j}+\hat{k}$,$\hat{i}-3\hat{j}-5\hat{k}$,and $a\hat{i}-3\hat{j}+\hat{k}$ respectively are the vertices of a right-angled triangle with $m\angle C = 90^\circ$ are:

Find the angle between the diagonals of parallelogram $PQRS$,if $\vec{PQ} = 3\hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{PS} = \hat{i} - 2\hat{k}$.