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

Let $O$ be the origin, $\vec{OP} = \vec{a}$ and $\vec{OQ} = \vec{b}$. If $R$ is the point on $\vec{OP}$ such that $\vec{OP} = 5\vec{OR}$, and $M$ is the point such that $\vec{OQ} = 5\vec{RM}$, then $\vec{PM}$ is equal to:

If $A \equiv (2i + 3j)$, $B \equiv (\pi + 9j)$, and $C \equiv (i - j)$ are collinear, then the value of $p$ is: (in $/2$)

The ratio in which $\hat{i}+2 \hat{j}+3 \hat{k}$ divides the join of $-2 \hat{i}+3 \hat{j}+5 \hat{k}$ and $7 \hat{i}-\hat{k}$ is

Let $\overline{OA}=\overline{a}, \overline{OB}=\overline{b}$. If the vector along the angle bisector of $\angle AOB$ is given by $x \frac{\overline{a}}{|\overline{a}|}+y \frac{\overline{b}}{|\overline{b}|}$,then which of the following is true?

Consider two points $P$ and $Q$ with position vectors $\overrightarrow{OP} = 3\vec{a} - 2\vec{b}$ and $\overrightarrow{OQ} = \vec{a} + \vec{b}$. Find the position vector of a point $R$ which divides the line joining $P$ and $Q$ in the ratio $2:1$ externally.