$ABCD$ is a parallelogram such that $L$ is the mid-point of $BC$. Then,$\vec{AL}$ is equal to:

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}$ be two vectors such that $\vec{a} \cdot \vec{b}=1$,$\cos(\theta) = \frac{1}{3}$ where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$,and the components of $\vec{b}$ with respect to $(\hat{i}, \hat{j}, \hat{k})$ are integers. Then the number of possible vectors that represent $\vec{b}$ is

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:

Let $u, v$ and $w$ be non-coplanar vectors. Then the points corresponding to which of the following vectors are collinear?

$P$ is a point on the side $BC$ of the $\Delta ABC$ and $Q$ is a point such that $\overrightarrow{PQ}$ is the resultant of $\overrightarrow{AP}, \overrightarrow{PB}, \overrightarrow{PC}$. Then $ABQC$ is a

The position vector of point $C$ relative to $B$ is $(\hat{i} + \hat{j})$ and the position vector of $B$ relative to $A$ is $(\hat{i} - \hat{j})$. The position vector of $C$ relative to $A$ is: