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:

Let $\vec{u} = 2 \hat{i} + \hat{j}$ and $\vec{v} = 3 \hat{i} - 5 \hat{j}$. Consider three points $P, Q,$ and $R$ having the position vectors $\left(\frac{5}{2}\right) \hat{i} - 2 \hat{j}, \left(\frac{7}{3}\right) \hat{i} - \hat{j},$ and $\left(\frac{9}{4}\right) \hat{i}$ respectively. Among these,the points on the line passing through $\vec{u}$ and $\vec{v}$ are

Let $\vec{a}, \vec{b}, \vec{c}$ be three coplanar concurrent vectors such that the angle between any two of them is the same. If the product of their magnitudes is $14$ and $(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) + (\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) + (\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b}) = 168$,then $|\vec{a}| + |\vec{b}| + |\vec{c}|$ is equal to:

If $\vec{a} = 4 \hat{i} + 5 \hat{j} - 3 \hat{k}$ and $\vec{b} = 6 \hat{i} - 2 \hat{j} - 2 \hat{k}$ are two vectors,then the magnitude of the component of $\vec{b}$ parallel to $\vec{a}$ is: (in $\sqrt{2}$)

The component of $\vec{i} + \vec{j}$ along $\vec{j} + \vec{k}$ is:

If $|a|=3, |b|=4$ and the angle between $a$ and $b$ is $120^{\circ}$,then $|4a+3b|$ is equal to