$A$ force $\overrightarrow{F} = 2i + j - k$ acts at a point $A$,whose position vector is $2i - j$. The moment of $\overrightarrow{F}$ about the origin is:

Find the unit vector perpendicular to the plane containing the vectors $\vec{a} = 2\hat{i} - 6\hat{j} - 3\hat{k}$ and $\vec{b} = 4\hat{i} + 3\hat{j} - \hat{k}$.

$A$ unit vector orthogonal to the vector $\vec{a} = 3\hat{i} + 4\hat{j} + 5\hat{k}$ and coplanar with the vectors $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{c} = \hat{i} - \hat{j} + \hat{k}$ is:

Let $\bar{a}=\alpha \hat{i}+3 \hat{j}-\hat{k}$,$\bar{b}=3 \hat{i}-\beta \hat{j}+4 \hat{k}$ and $\overline{c}=\hat{i}+2 \hat{j}-2 \hat{k}$,where $\alpha, \beta \in R$,be three vectors. If the projection of $\overline{a}$ on $\overline{c}$ is $\frac{10}{3}$ and $\bar{b} \times \bar{c}=-6 \hat{i}+10 \hat{j}+7 \hat{k}$,then the value of $2 \alpha+\beta$ 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

If $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}$, $\vec{b}=3(\hat{i}-\hat{j}+\hat{k})$ and $\vec{c}$ is a vector such that $\vec{a} \times \vec{c}=\vec{b}$ and $\vec{a} \cdot \vec{c}=3$, then $\vec{a} \cdot(\vec{c} \times \vec{b}-\vec{b}-\vec{c})=$