Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a}=\hat{i}-\hat{j}+3\hat{k}$ and $\vec{b}=2\hat{i}-7\hat{j}+\hat{k}$.

Let $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ be three unit vectors such that $\vec{\alpha} \cdot \vec{\beta} = \vec{\alpha} \cdot \vec{\gamma} = 0$ and the angle between $\vec{\beta}$ and $\vec{\gamma}$ is $30^{\circ}$. Then $\vec{\alpha}$ is

Let $A(2, 3, 5)$ and $C(-3, 4, -2)$ be opposite vertices of a parallelogram $ABCD$. If the diagonal $\overrightarrow{BD} = \hat{i} + 2\hat{j} + 3\hat{k}$,then the area of the parallelogram is equal to:

The position vectors of the points $A, B, C$ are $\hat{i}+2\hat{j}-\hat{k}, \hat{i}+\hat{j}+\hat{k}$,and $2\hat{i}+3\hat{j}+2\hat{k}$ respectively. If $A$ is chosen as the origin,then the cross product of the position vectors of $B$ and $C$ is:

For any three vectors $\vec{a}, \vec{b}, \vec{c}$,the value of $\vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times (\vec{c} + \vec{a}) + \vec{c} \times (\vec{a} + \vec{b})$ is equal to:

The diagonals of a parallelogram are $\vec{d_1} = \hat{j} + \hat{k}$ and $\vec{d_2} = \hat{i} + \hat{j}$. The area of the parallelogram is . . . . . . sq. units.