If $a$ and $b$ are two non-collinear vectors and $xa + yb = 0$,then:

Find the unit vector in the direction of the sum of the vectors,$\vec{a}=2 \hat{i}+2 \hat{j}-5 \hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}+3 \hat{k}$.

If $a, b$ and $c$ are three non-collinear points and $ka + 2b + 3c$ is a point in the plane of $a, b$ and $c$, then $k =$

If $\vec{a}=4 \hat{i}+13 \hat{j}-18 \hat{k}$,$\vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}$,and $\vec{c}=2 \hat{i}+3 \hat{j}-4 \hat{k}$ are three vectors such that $\vec{a}=x \vec{b}+y \vec{c}$,then $x+y=$

$A$ vector $\vec{r}$ of magnitude $3\sqrt{2}$ units which makes angles of $\pi/4$ and $\pi/2$ respectively with $Y$ and $Z$ axes is:

If $|\overline{a}|=2, |\overline{b}|=3, |\overline{c}|=5$ and each of the angles between the vectors $\overline{a}$ and $\overline{b}$,$\overline{b}$ and $\overline{c}$,and $\overline{c}$ and $\overline{a}$ is $60^{\circ}$,then the value of $|\overline{a}+\overline{b}+\overline{c}|$ is