Let $\vec{p}$ and $\vec{q}$ be the position vectors of points $P$ and $Q$ respectively,with respect to the origin $O$,and let $|\vec{p}|=p, |\vec{q}|=q$. The points $R$ and $S$ divide the line segment $PQ$ internally and externally in the ratio $2:3$ respectively. If $\vec{OR}$ and $\vec{OS}$ are perpendicular,then:

$\bar{a}, \bar{b}, \bar{c}$ are nonzero vectors such that $\bar{a}$ is perpendicular to $\bar{b}$ and $\bar{c}$,$|\bar{a}|=1, |\bar{b}|=2, |\bar{c}|=1$ and $\bar{b} \cdot \bar{c}=1$. There is a nonzero vector $\bar{d}$ coplanar with $\bar{a}+\bar{b}$ and $2\bar{b}-\bar{c}$. If $\bar{d} \cdot \bar{a}=1$,then $|\bar{d}|^2=$ (Note that $x$ and $y$ are parameters involved when we write $\bar{d}=x(\bar{a}+\bar{b})+y(2\bar{b}-\bar{c})$)

If $\overline{a} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$ and $\overline{b} = \hat{i} - \hat{j} - \hat{k}$,then the projection of $\overline{b}$ in the direction of $\overline{a}$ is

Find $|\vec{x}|$,if for a unit vector $\vec{a}$,$(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a}) = 12$.

Given the following simultaneous equations for vectors $x$ and $y$:
$(i) x + y = a$
$(ii) x \times y = b$
$(iii) x \cdot a = 1$
Then $x = ?, y = ?$

If $a, b$ and $c$ are perpendicular to $b + c, c + a$ and $a + b$ respectively,and if $|a + b| = 6, |b + c| = 8$ and $|c + a| = 10$,then $|a + b + c| = $