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

If the points $P(1, -1, 2)$,$Q(2, 0, -1)$,and $R(0, 2, 1)$ are coplanar,find the unit vector perpendicular to the plane containing these points.

The position vectors of the points $A, B, C$ are $\hat{i}+2\hat{j}-\hat{k}, \hat{i}+\hat{j}+\hat{k}$,and $2\hat{i}+3\hat{j}+2\hat{k}$ respectively. If $A$ is chosen as the origin,then the cross product of the position vectors of $B$ and $C$ is:

Let $\vec{a}=2 \hat{i}+\alpha \hat{j}+\hat{k}$,$\vec{b}=-\hat{i}+\hat{k}$,and $\vec{c}=\beta \hat{j}-\hat{k}$,where $\alpha$ and $\beta$ are integers and $\alpha \beta=-6$. Let the values of the ordered pair $(\alpha, \beta)$ for which the area of the parallelogram with diagonals $\vec{a}+\vec{b}$ and $\vec{b}+\vec{c}$ is $\frac{\sqrt{21}}{2}$,be $(\alpha_1, \beta_1)$ and $(\alpha_2, \beta_2)$. Then $\alpha_1^2+\beta_1^2-\alpha_2 \beta_2$ is equal to

Let $\overrightarrow{a}=3\hat{i}-\hat{j}+2\hat{k}$,$\overrightarrow{b}=\overrightarrow{a}\times(\hat{i}-2\hat{k})$ and $\overrightarrow{c}=\overrightarrow{b}\times\hat{k}$. Then the projection of $\overrightarrow{c}-2\hat{j}$ on $\overrightarrow{a}$ is:

If $\vec{u} = \vec{a} - \vec{b}$ and $\vec{v} = \vec{a} + \vec{b}$ and $|\vec{a}| = |\vec{b}| = 2$,then $|\vec{u} \times \vec{v}| = ......$