The position vectors of the points $A$ and $B$ with respect to $O$ are $2 \hat{i}+2 \hat{j}+\hat{k}$ and $2 \hat{i}+4 \hat{j}+4 \hat{k}$. The length of the internal bisector of $\angle BOA$ of $\triangle AOB$ is:

Given three vectors $\bar{a}, \bar{b}, \bar{c}$,two of which are collinear. If $\bar{a}+\bar{b}$ is collinear with $\bar{c}$ and $\bar{b}+\bar{c}$ is collinear with $\bar{a}$,and $|\bar{a}|=|\bar{b}|=|\bar{c}|=\sqrt{2}$,then $\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a}=$

Let $\vec{a}=4 \hat{i}-\hat{j}+\hat{k}$,$\vec{b}=11 \hat{i}-\hat{j}+\hat{k}$,and $\vec{c}$ be a vector such that $(\vec{a}+\vec{b}) \times \vec{c} = \vec{c} \times (-2 \vec{a}+3 \vec{b})$. If $(2 \vec{a}+3 \vec{b}) \cdot \vec{c} = 1670$,then $|\vec{c}|^2$ is equal to:

An arc $PQ$ of a circle subtends a right angle at its centre $O$. The midpoint of the arc $PQ$ is $R$. If $\vec{OP}=\vec{u}$,$\vec{OR}=\vec{v}$ and $\vec{OQ}=\alpha \vec{u}+\beta \vec{v}$,then $\alpha, \beta^2$ are the roots of the equation

The magnitude of the projection of the vector $\vec{a} = 4\hat{i} - 3\hat{j} + 2\hat{k}$ on the line which makes equal angles with the coordinate axes is

Let $\vec{u}=\hat{i}-\hat{j}-2\hat{k}$,$\vec{v}=2\hat{i}+\hat{j}-\hat{k}$,$\vec{v} \cdot \vec{w}=2$ and $\vec{v} \times \vec{w}=\vec{u}+\lambda\vec{v}$. Then $\vec{u} \cdot \vec{w}$ is equal to $......$