$ABCDE$ is a pentagon. Forces $\overline{AB}, \overline{AE}, \overline{DC}, \overline{ED}$ act at a point. Which force should be added to this system to make the resultant $2\overline{AC}$?

If $a = \hat{i} + 2 \hat{j} + 3 \hat{k}$,$b = 2 \hat{i} + 3 \hat{j} + \hat{k}$,$c = 8 \hat{i} + 13 \hat{j} + 9 \hat{k}$ and $x a + y b + z c = 0$,then $\frac{x y}{z^2} =$

If $\bar{a}=2 \hat{i}-\hat{j}+\hat{k}$,$\bar{b}=\hat{i}+\hat{j}-2 \hat{k}$ and $\bar{c}=4 \hat{i}-2 \hat{j}+\hat{k}$,then the unit vector in the direction of $3 \bar{a}+\bar{b}-2 \bar{c}$ is

Let $\bar{a}, \bar{b}$ and $\bar{c}$ be non-coplanar vectors. If $P, Q, R$ and $S$ are four points with position vectors $-\bar{a}+4\bar{b}-3\bar{c}$,$3\bar{a}+2\bar{b}-5\bar{c}$,$-3\bar{a}+8\bar{b}-5\bar{c}$ and $-3\bar{a}+2\bar{b}+\bar{c}$ respectively,then the ordered pair $(x, y)$ of real numbers such that $\overline{PQ} = x \cdot \overline{PR} + y \cdot \overline{PS}$ is

If $ABCD$ is a parallelogram and the position vectors of $A, B, C$ are $i + 3j + 5k, i + j + k$ and $7i + 7j + 7k$, then the position vector of $D$ will be:

Given vectors $\vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} - \hat{k}$. The vector in the direction of $\vec{a} + \vec{b}$ with magnitude $\sqrt{2}$ is . . . . . . .