The vectors $\vec{AB} = 3\hat{i} + 4\hat{k}$ and $\vec{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}$ are the sides of a $\triangle ABC$. The length of the median through $A$ is:

If the position vectors of two points $A$ and $B$ are $\vec{a} + 3\vec{b}$ and $\vec{a} - 2\vec{b}$ respectively,find the position vector of the point that divides $AB$ in the ratio $2:5$.

The direction cosines of the vector $\hat{i}-2\hat{j}+3\hat{k}$ are . . . . . . .

If a point $C$ divides the line segment joining the points with the position vectors $2 \hat{i}-3 \hat{j}+2 \hat{k}$ and $3 \hat{i}-\hat{j}-2 \hat{k}$ in the ratio $2: 3$,then the distance of $C$ from the point with position vector $2 \hat{i}-\hat{j}+\hat{k}$ is

The magnitudes of mutually perpendicular forces $a, b$ and $c$ are $2, 10$ and $11$ respectively. Then the magnitude of its resultant is

The direction cosines of the vector $\vec{a} = -2 \hat{i} + \hat{j} - 5 \hat{k}$ are