$A$ tetrahedron has vertices $O(0,0,0)$,$A(1,2,1)$,$B(2,1,3)$,and $C(-1,1,2)$. The angle between the faces $OAB$ and $ABC$ is:

The number of vectors of unit length perpendicular to the two vectors $a=(1,1,0)$ and $b=(0,1,1)$ is

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.

Let $\vec{a} = 3\hat{i} + 2\hat{j} + x\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$,for some real $x$. Then $|\vec{a} \times \vec{b}| = r$ is possible if

Let $\vec \alpha = 3\hat i + \hat j$ and $\vec \beta = 2\hat i - \hat j + 3\hat k.$ If $\vec \beta = \vec \beta _1 - \vec \beta _2,$ where $\vec \beta _1$ is parallel to $\vec \alpha$ and $\vec \beta _2$ is perpendicular to $\vec \alpha,$ then $\vec \beta _1 \times \vec \beta _2$ is equal to

If $\bar{a}, \bar{b}, \bar{c}$ are three coplanar vectors such that $|\bar{a}|=1, |\bar{b}|=2$,$\bar{b} \cdot \bar{c}=8$,and the angle between $\bar{b}$ and $\bar{c}$ is $45^{\circ}$,then $|\bar{a} \times (\bar{b} \times \bar{c})|=$