If a vector $a$ of magnitude $50$ is collinear with the vector $b = 6i - 8j - \frac{15}{2}k$, and makes an acute angle with the positive direction of the $z$-axis, then the vector $a$ is equal to:

In a regular hexagon $ABCDEF$,$\overrightarrow{AE} = $

$2 \hat{i}-3 \hat{j}+\hat{k}$ and $\hat{i}+2 \hat{j}-3 \hat{k}$ are the position vectors of two points $A$ and $B$ respectively and $C$ divides $AB$ in the ratio $3:2$. If $3 \hat{i}-\hat{j}+2 \hat{k}$ is the position vector of a point $D$,then the unit vector in the direction of $\overrightarrow{CD}$ is

If $\overline{a} = m \overline{b} + n \overline{c}$,where $\overline{a} = 4 \hat{i} + 13 \hat{j} - 18 \hat{k}$,$\overline{b} = \hat{i} - 2 \hat{j} + 3 \hat{k}$,and $\overline{c} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$,then $m + n =$

$(a \cdot i)i + (a \cdot j)j + (a \cdot k)k = $

The value of $\hat{k} \cdot (\hat{i} \times \hat{j}) + \hat{i} \cdot (\hat{j} \times \hat{k})$ is . . . . . . .