If $\theta$ is the angle between the unit vectors $a$ and $b$,then $\sin \frac{\theta}{2}$ is equal to

Let $a, b, c$ be the position vectors of the vertices of a triangle $ABC$. The vector area of triangle $ABC$ is

Let $\vec{a}, \vec{b}, \vec{c}$ be non-coplanar vectors. If the three points with position vectors $\lambda \vec{a}-2 \vec{b}+\vec{c}$,$2 \vec{a}+\lambda \vec{b}-2 \vec{c}$,and $4 \vec{a}+7 \vec{b}-8 \vec{c}$ are collinear,then $\lambda=$

Let $\overline{A}=2 \hat{i}+\hat{k}$,$\overline{B}=\hat{i}+\hat{j}+\hat{k}$ and $\overline{C}=4 \hat{i}-3 \hat{j}+7 \hat{k}$. If a vector $\overline{R}$ satisfies $\overline{R} \times \overline{B}=\overline{C} \times \overline{B}$ and $\overline{R} \cdot \overline{A}=0$,then $\overline{R}$ is given by

$ABCD$ is a quadrilateral with $\overline{AB}=\bar{a}$,$\overline{AD}=\bar{b}$ and $\overline{AC}=2\bar{a}+3\bar{b}$. If its area is $\alpha$ times the area of the parallelogram with $AB$ and $AD$ as adjacent sides,then the value of $\alpha$ is

$A$ particle is acted upon by constant forces $4\hat{i} + \hat{j} - 3\hat{k}$ and $3\hat{i} + \hat{j} - \hat{k}$. The displacement of the particle from the point $\hat{i} + 2\hat{j} + 3\hat{k}$ to the point $5\hat{i} + 4\hat{j} + \hat{k}$ is given. Find the total work done by the forces in units.