The perimeter of the triangle whose vertices have the position vectors $(i + j + k)$,$(5i + 3j - 3k)$,and $(2i + 5j + 9k)$ is given by:

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:

Let $ABC$ be a triangle and $\bar{a}, \bar{b}, \bar{c}$ be the position vectors of $A, B, C$ respectively. Let $D$ divide $BC$ in the ratio $3:1$ internally and $E$ divide $AD$ in the ratio $4:1$ internally. Let $BE$ meet $AC$ in $F$. If $E$ divides $BF$ in the ratio $3:2$ internally, then the position vector of $F$ is

In a parallelogram $ABCD$,if the diagonals are $\overline{AC}$ and $\overline{BD}$,then $\overline{AC} - \overline{BD} = \dots$

If $\theta$ is the interior angle of a regular pentagon,then $|(\sin \theta) \hat{i}+(\cos \theta) \hat{j}+(\tan \theta) \hat{k}|=$

$A$ vector of magnitude $\sqrt{2}$ units along the internal bisector of the angle between the vectors $\vec{a} = 2 \hat{i} - 2 \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ is