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.

For some real number $\lambda$,if the area of the triangle having $\vec{a}=3 \hat{i}-\hat{j}+\lambda \hat{k}$ and $\vec{b}=\lambda \hat{i}+\hat{j}-3 \hat{k}$ as two of its sides is $\frac{\sqrt{195}}{2}$,then the number of distinct possible values of $\lambda$ is

Let $L_1: \frac{x+1}{3}=\frac{y+2}{2}=\frac{z+1}{1}$ and $L_2: \frac{x-2}{2}=\frac{y+2}{1}=\frac{z-3}{3}$ be the given lines. Then the unit vector perpendicular to $L_1$ and $L_2$ is

If $\vec{a} = 2 \hat{i} + 2 \hat{j} + \hat{k}$,$|\vec{b}| = 6$ and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$,then the area of the triangle (in square units) with $\vec{a}$ and $\vec{b}$ as two of its sides is

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

Let $\bar{a}, \bar{b}$ and $\bar{c}$ be three unit vectors such that $\bar{a} \times(\bar{b} \times \bar{c})=\frac{\sqrt{3}}{2}(\bar{b}+\bar{c})$. If $\bar{b}$ is not parallel to $\bar{c}$,then the angle between $\bar{a}$ and $\bar{b}$ is