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:

$A$ vector of magnitude $\sqrt{51}$ which makes equal angles with the vectors $\bar{a}=\frac{1}{3}(\bar{i}-2 \bar{j}+2 \bar{k})$,$\bar{b}=\frac{1}{5}(-4 \bar{i}-3 \bar{k})$ and $\bar{c}=\bar{j}$,is

The angle between the vectors $\vec{a} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$ is

If $\vec{a}$ is a unit vector and $(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a})=8$,then find $|\vec{x}|$.

If the position vectors of $A, B$ and $C$ are respectively $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $3 \hat{i}-4 \hat{j}-4 \hat{k}$,then $\cos ^2 A$ is equal to

Let the vectors $\vec{a} = -\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} + \hat{k}$. For some $\lambda, \mu \in \mathbb{R}$, let $\vec{c} = \lambda \vec{a} + \mu \vec{b}$. If $\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10$ and $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2$, then $|\vec{c}|^2$ is equal to: