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

Find a vector that is perpendicular to both vectors $\hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + \hat{j}$.

If the area of the triangle with vertices $\hat{i}+y \hat{j}$,$\hat{i}+2 \hat{k}$,and $3 \hat{j}+\hat{k}$ is $\sqrt{6}$ sq. units,then the values of $y$ are

Let $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ be the position vectors of the vertices $A, B, C$ respectively of $\triangle ABC$. The vector area of $\triangle ABC$ is:

Let $x$ and $y$ be real numbers. If $\vec{a}=(\sin x) \hat{i}+(\sin y) \hat{j}$ and $\vec{b}=(\cos x) \hat{i}+(\cos y) \hat{j}$,then $|\vec{a} \times \vec{b}|$ is

If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$, $\overrightarrow{b}=\hat{i}+\hat{j}$, $\overrightarrow{c}=\hat{i}$ and $(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}$, then $\lambda+\mu$ is equal to: