If $D, E$ and $F$ are the mid-points of the sides $BC, CA$ and $AB$ of triangle $ABC$ respectively,then $\overline{AD} + \frac{2}{3} \overline{BE} + \frac{1}{3} \overline{CF} =$

If $\vec{a}$ and $\vec{b}$ are non-collinear vectors,then the value of $\alpha$ for which the vectors $\vec{u} = (\alpha - 2)\vec{a} + \vec{b}$ and $\vec{v} = (2 + 3\alpha)\vec{a} - 3\vec{b}$ are collinear is:

If $\vec{a} = 7\hat{j} + 10\hat{k}$, $\vec{b} = -\hat{i} + 6\hat{j} + 6\hat{k}$ and $\vec{c} = -4\hat{i} + 9\hat{j} + 6\hat{k}$ are the position vectors of the vertices $A, B$ and $C$ respectively of $\triangle ABC$. Then the position vector of the point where the bisector of the angle $A$ meets side $BC$ is

If $\vec{a}=\vec{b}+\vec{c},$ then is it true that $|\vec{a}|=|\vec{b}|+|\vec{c}|$ ? Justify your answer.

The position vectors of the points $A, B, C$ are $(2\hat{i}+\hat{j}-\hat{k}), (3\hat{i}-2\hat{j}+\hat{k})$ and $(\hat{i}+4\hat{j}-3\hat{k})$ respectively. These points

If the vectors $3i - 2j + 5k$ and $-2i + pj - qk$ are collinear,then $(p, q) = $?