Let $a$ and $b$ be two unit vectors inclined at an angle $\theta$,then $\sin(\theta/2)$ is equal to

Let $\vec{a} = 2\hat{i} + \hat{k}$, $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{c} = 4\hat{i} - 3\hat{j} + 7\hat{k}$. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$ and $\vec{r} \cdot \vec{a} = 0$, then $\vec{r} \cdot \vec{c} = $

If the area of a parallelogram whose diagonals are represented by vectors $\vec{d_1} = 3 \hat{i} + \lambda \hat{j} + 2 \hat{k}$ and $\vec{d_2} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ is $\frac{\sqrt{117}}{2}$ sq. units,then $\lambda=$

$A$ vector whose modulus is $\sqrt{51}$ and makes the same angle with $a = \frac{i - 2j + 2k}{3}$,$b = \frac{-4i - 3k}{5}$,and $c = j$ is:

In a parallelogram $OACB$,$\overrightarrow{OA} = \vec{a}$,$\overrightarrow{OB} = \vec{b}$,and the foot of the perpendicular drawn from point $B$ to $AC$ is $M$. If $\vec{a} \cdot \vec{b} = 1$ and $|\vec{a}| = |\vec{b}| = 2$,then $|\overrightarrow{BM}|$ is:

If the adjacent sides of a rectangle are $\bar{a}=5\bar{m}-3\bar{n}$,$\bar{b}=-\bar{m}-2\bar{n}$ and the adjacent sides of another rectangle are $\bar{c}=-4\bar{m}-\bar{n}$,$\bar{d}=-\bar{m}+\bar{n}$,then the angle between the vectors $\bar{x}=\frac{\bar{a}+\bar{c}+\bar{d}}{3}$ and $\bar{y}=\frac{\bar{c}+\bar{d}}{5}$ is