If the vectors $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$,$\vec{b} = 2\hat{i} + 4\hat{j} + \hat{k}$,and $\vec{c} = \alpha\hat{i} + \hat{j} + \beta\hat{k}$ are mutually orthogonal,then $(\alpha, \beta) = $

If the vector $a = 3\hat{j} + 4\hat{k}$ is the sum of two vectors $a_1$ and $a_2$, where vector $a_1$ is parallel to $b = \hat{i} + \hat{j}$ and vector $a_2$ is perpendicular to $b$, then $a_1 =$

Let $\vec{a}, \vec{b}, \vec{c}$ be three unit vectors such that $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} - \vec{a} \cdot \vec{c} = \frac{3}{2}$. Then the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$ is:

Let $ABC$ be a triangle. Let a point $P$ divide $AB$ in the ratio $1:2$ internally and a point $Q$ divide $BC$ in the ratio $1:2$ internally. Let $D$ be the point of intersection of $AQ$ and $CP$. If the area of the triangle $ABC$ is $k$ square units,then the area of the triangle $BCD$ in square units is:

In a parallelogram $ABCD$,$|\overline{AB}| = a$,$|\overline{AD}| = b$ and $|\overline{AC}| = c$,then the value of $\overline{DA} \cdot \overline{AB}$ is:

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