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

If the vectors $\vec{a} = \hat{i} - 3\hat{j} + 2\hat{k}$ and $\vec{b} = -\hat{i} + 2\hat{j}$ represent the diagonals of a parallelogram,then its area will be:

Let $\vec{b}=\hat{i}+\hat{j}+\lambda \hat{k}, \lambda \in R$. If $\vec{a}$ is a vector such that $\vec{a} \times \vec{b}=13 \hat{i}-\hat{j}-4 \hat{k}$ and $\vec{a} \cdot \vec{b}+21=0$,then $(\vec{b}-\vec{a}) \cdot(\hat{k}-\hat{j})+(\vec{b}+\vec{a}) \cdot(\hat{i}-\hat{k})$ is equal to

The area (in sq. units) of the parallelogram whose diagonals are along the vectors $8\hat{i} - 6\hat{j}$ and $3\hat{i} + 4\hat{j} - 12\hat{k}$ is:

If $\theta$ is the angle between the vectors $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + \hat{k}$,find the value of $\sin \theta$.

Let $\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\bar{b}=\hat{i}+\hat{j}$. 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 $|(\bar{a} \times \bar{b}) \times \bar{c}|=$