If $\vec{\alpha} = 3\hat{i} - \hat{k}$, $|\vec{\beta}| = \sqrt{5}$, and $\vec{\alpha} \cdot \vec{\beta} = 3$, then the area of the parallelogram for which $\vec{\alpha}$ and $\vec{\beta}$ are adjacent sides is:

If the position vectors of the vertices $A, B, C$ of a triangle $ABC$ are $4 \hat{\imath} + 7 \hat{\jmath} + 8 \hat{k}$,$2 \hat{\imath} + 3 \hat{\jmath} + 4 \hat{k}$,and $2 \hat{\imath} + 5 \hat{\jmath} + 7 \hat{k}$ respectively,then the position vector of the point where the bisector of angle $A$ meets $BC$ is

Three non-coplanar vectors $\bar{a}, \bar{b}, \bar{c}$ are the coterminous edges of a parallelepiped. If $\bar{a}$ and $\bar{b}$ determine the base of the parallelepiped, then its height is:

Given the following simultaneous equations for vectors $x$ and $y$:
$(i) x + y = a$
$(ii) x \times y = b$
$(iii) x \cdot a = 1$
Then $x = ?, y = ?$

If $\vec{a} = -4 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and $\vec{b} = \sqrt{2} \hat{i} - \sqrt{2} \hat{j}$ are two vectors,then the angle between the vectors $2 \vec{a}$ and $\frac{\vec{b}}{2}$ is (in $^{\circ}$)

If the vectors $\bar{a}, \bar{b}, \bar{c}$ satisfy the condition $|\bar{a}-\bar{c}|=|\bar{b}-\bar{c}|$,then $(\bar{b}-\bar{a}) \cdot \left(\bar{c}-\frac{\bar{a}+\bar{b}}{2}\right) = $