If the vectors $a\,i - 2j + 3k$ and $3i + 6j - 5k$ are perpendicular to each other,then $a$ is given by

For any two non-zero vectors $\vec{a}$ and $\vec{b}$,$(a \vec{b} + b \vec{a}) \cdot (a \vec{b} - b \vec{a})$ is equal to:

Let two non-collinear vectors $\hat{a}$ and $\hat{b}$ form an acute angle. $A$ point $P$ moves such that at any time $t$,the position vector $\overline{OP}$,where $O$ is the origin,is given by $\hat{a} \sin t + \hat{b} \cos t$. When $P$ is farthest from the origin $O$,let $M$ be the length of $\overline{OP}$ and $\hat{u}$ be the unit vector along $\overline{OP}$. Then:

Consider the vectors $\vec{a}=3 \hat{i}+5 \hat{j}+2 \hat{k}$, $\vec{b}=2 \hat{i}-3 \hat{j}-5 \hat{k}$ and $\vec{c}=-5 \hat{i}-2 \hat{j}+3 \hat{k}$. If $l, m$ and $n$ are the lengths of the projections of $\vec{a}$ on $\vec{b}$, $\vec{b}$ on $\vec{c}$ and $\vec{c}$ on $\vec{a}$ respectively, then:

If $\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=3 \hat{i}+\hat{j}$ are such that $\vec{a}+\lambda \vec{b}$ is perpendicular to $\vec{c},$ then find the value of $\lambda$.

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: