$A$ particle acted on by constant forces $4i + j - 3k$ and $3i + j - k$ is displaced from the point $i + 2j + 3k$ to the point $5i + 4j + k$. The total work done by the force is ............... $unit$.

Let $\overline{a}=3 \hat{i}-\alpha \hat{j}+\hat{k}$ and $\overline{b}=\hat{i}+\alpha \hat{j}+3 \hat{k}$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\overline{a}$ and $\overline{b}$ is $8 \sqrt{3}$ sq. units,then $\overline{a} \cdot \overline{b}$ is equal to

The value of $b$ such that the scalar product of the vector $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ with the unit vector parallel to the sum of the vectors $\vec{u} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\vec{v} = b\hat{i} + 2\hat{j} + 3\hat{k}$ is $1$, is...

If $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$ and $|\overrightarrow{a}|=3, |\overrightarrow{b}|=4$ and $|\overrightarrow{c}|=\sqrt{37}$, then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is:

If $|\vec{a}|=4, |\vec{b}|=5, |\vec{a}-\vec{b}|=3$ and $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$, then $\cot^2 \theta=$

Let $\vec{c}$ be the projection vector of $\vec{b}=\lambda \hat{i}+4 \hat{k}, \lambda>0$,on the vector $\vec{a}=\hat{i}+2 \hat{j}+2 \hat{k}$. If $|\vec{a}+\vec{c}|=7$,then the area of the parallelogram formed by the vectors $\vec{b}$ and $\vec{c}$ is . . . . . . .