The area of a parallelogram whose adjacent sides are $\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b} = -\hat{j} - 2\hat{k}$ is . . . . . . sq. units.

Find the magnitude of the torque of a couple formed by a force $\vec{F} = 3\hat{i} + 2\hat{j} - \hat{k}$ acting at the point $\hat{i} - \hat{j} + \hat{k}$ and the force $-\vec{F}$ acting at the point $2\hat{i} - 3\hat{j} - \hat{k}$.

Let $\bar{a} = \hat{i} + 2\hat{j} - 2\hat{k}$ and $\bar{b} = \hat{i} - \hat{j} + \hat{k}$. If $\bar{c}$ is a vector such that $\bar{a} \cdot \bar{c} = |\bar{c}|$,$|\bar{c} - \bar{a}| = 2\sqrt{2}$ and the angle between $\bar{a} \times \bar{b}$ and $\bar{c}$ is $60^{\circ}$,then $|(\bar{a} \times \bar{b}) \times \bar{c}|$ is equal to

If $\bar{a}=2\hat{i}+3\hat{j}-\hat{k}$,$\bar{b}=-\hat{i}+2\hat{j}-4\hat{k}$ and $\bar{c}=\hat{i}+\hat{j}+\hat{k}$,then $(\bar{a} \times \bar{b}) \cdot(\bar{a} \times \bar{c})=$

If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of the vertices of a triangle,show that $\frac{1}{2}[\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}]$ gives the vector area of the triangle. Hence,deduce the condition that the three points $\vec{a}, \vec{b}, \vec{c}$ are collinear. Also,find the unit vector normal to the plane of the triangle.

If the area of the parallelogram with $\bar{a}$ and $\bar{b}$ as two adjacent sides is $20$ square units,then the area of the parallelogram having $3 \bar{a} + \bar{b}$ and $2 \bar{a} + 3 \bar{b}$ as two adjacent sides in square units is