Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}|=\sqrt{14}$,$|\vec{b}|=\sqrt{6}$ and $|\vec{a} \times \vec{b}|=\sqrt{48}$. Then $(\vec{a} \cdot \vec{b})^2$ is equal to $...........$.

If $|\vec{a}| = 3, |\vec{b}| = 4, |\vec{c}| = 5$ such that each vector is perpendicular to the sum of the other two, then $|\vec{a} + \vec{b} + \vec{c}|$ is equal to

Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of the vertices of a triangle $ABC$. Through the vertices,lines are drawn parallel to the sides to form the triangle $A'B'C'$. Then the centroid of $\Delta A'B'C'$ is

If $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$,then find $\vec{a} \cdot \vec{b}$.

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = \hat{i} - \hat{j} + \hat{k}$,and $\vec{c} = \hat{i} + \hat{j} - \hat{k}$ be three vectors. If $\vec{v}$ is a vector in the plane of $\vec{a}$ and $\vec{b}$ such that the projection of $\vec{v}$ on $\vec{c}$ is $\frac{1}{\sqrt{3}}$,then $\vec{v} = $

Find the angle $\theta$ between the vectors $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.