If $PQRST$ is a pentagon,then the resultant of forces $\overline{PQ}, \overline{PT}, \overline{QR}, \overline{SR}, \overline{TS}$ and $\overline{PS}$ is

In triangle $ABC$ (Fig.),which of the following is not true:

The components of a vector $\vec{a}$ with respect to a rectangular Cartesian system are $2p$ and $1$. The system is rotated about the origin through a certain angle in the counter-clockwise direction. If the components of $\vec{a}$ with respect to the new system are $p + 1$ and $1$,then:

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

Let $A, B, C, D$ be the points in the plane with position vectors $\vec{a} = -2\hat{i} - \hat{j}$, $\vec{b} = 4\hat{i}$, $\vec{c} = 3\hat{i} + 3\hat{j}$, and $\vec{d} = -3\hat{i} + 2\hat{j}$ respectively. Then $ABCD$ is:

Using vectors,find the value of $k$ such that the points $(k,-10,3), (1,-1,3)$ and $(3,5,3)$ are collinear.