If $ABCDEFGH$ is a convex octagon,then $\vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} + \vec{AH} + \vec{HG} + \vec{GF} + \vec{FE} = $

Let $PQRS$ be a quadrilateral. If $M$ and $N$ are midpoints of the sides $PQ$ and $RS$ respectively,then $\overrightarrow{PS} + \overrightarrow{QR} =$

If the position vectors of the vertices $A, B, C$ of a triangle $ABC$ are $7j + 10k$,$-i + 6j + 6k$,and $-4i + 9j + 6k$ respectively,then the triangle is:

If the vector $a = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ and $b$ are collinear and $|b| = 21$,then $b$ is equal to:

$A$ vector $\bar{a}$ has components $1$ and $2p$ with respect to a rectangular Cartesian system. This system is rotated through a certain angle about the origin in the counter-clockwise sense. If,with respect to the new system,$\bar{a}$ has components $1$ and $(p+1)$,then:

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