Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{b}|=1$ and $|\vec{b} \times \vec{a}|=2$. Then $|(\vec{b} \times \vec{a})-\vec{b}|^2$ is equal to

Let $A(2, 3, 0)$, $B(0, 3, 2)$ and $C(4, 0, 3)$ be the vertices of a triangle. Find the area of the triangle.

Let $\overrightarrow{a}$ be a non-zero vector parallel to the line of intersection of the two planes passing through the origin and containing the vectors $(\hat{i}+\hat{j}, \hat{i}+\hat{k})$ and $(\hat{i}-\hat{j}, \hat{j}-\hat{k})$ respectively. If $\theta$ is the angle between the vector $\vec{a}$ and the vector $\vec{b}=2\hat{i}-2\hat{j}+\hat{k}$ and $\vec{a} \cdot \vec{b}=6$,then the ordered pair $(\theta, |\vec{a} \times \vec{b}|)$ is equal to

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

If $\vec{a}=\hat{i}+\hat{j}-\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}$ is a unit vector perpendicular to $\vec{a}$ and coplanar with $\vec{a}$ and $\vec{b}$, then the unit vector $\vec{d}$ perpendicular to both $\vec{a}$ and $\vec{c}$ is

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