If the force $\overrightarrow{F} = \hat{i} + 2\hat{j} + 3\hat{k}$ moves a particle from position $\vec{r_1} = \hat{i} + \hat{j} - \hat{k}$ to $\vec{r_2} = 2\hat{i} - \hat{j} + \hat{k},$ then the work done is:

If the vectors $\hat{i}-2x\hat{j}-3y\hat{k}$ and $\hat{i}+3x\hat{j}+2y\hat{k}$ are orthogonal to each other,then the locus of the point $(x, y)$ is

If $\vec{a}$ and $\vec{b}$ are unit vectors such that the vector $\vec{a} + 3\vec{b}$ is perpendicular to $7\vec{a} - 5\vec{b}$,then find the angle between $\vec{a}$ and $\vec{b}$.

Let $\overline{u}, \overline{v}, \overline{w}$ be vectors such that $|\overline{u}|=1, |\overline{v}|=2, |\overline{w}|=3$. If the projection of $\overline{v}$ along $\overline{u}$ is equal to that of $\overline{w}$ along $\overline{u}$,and the vectors $\overline{v}$ and $\overline{w}$ are perpendicular to each other,then $|\overline{u}-\overline{v}+\overline{w}|$ equals:

Let $\hat{u}$ and $\hat{v}$ be unit vectors inclined at an acute angle such that $|\hat{u} \times \hat{v}| = \frac{\sqrt{3}}{2}$. If $\vec{A} = \lambda \hat{u} + \hat{v} + (\hat{u} \times \hat{v})$, then $\lambda$ is equal to:

If $c = 2 \lambda (a \times b) + 3 \mu (b \times a)$ where $a \times b \neq 0$ and $c \cdot (a \times b) = 0$,then: