If $a, b$ and $c$ are unit vectors such that $a+b+c=0$,then $a \cdot b+b \cdot c+c \cdot a$ is equal to

If $\bar{a}, \bar{b}, \bar{c}, \bar{d}$ are the position vectors of the points $A, B, C, D$ respectively such that $3 \bar{a}-\bar{b}+2 \bar{c}-4 \bar{d}=\overline{0}$,then the position vector of the point of intersection of the line segments $AC$ and $BD$ is

Let $\vec{a}, \vec{b}, \vec{c}$ be three unit vectors such that $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} - \vec{a} \cdot \vec{c} = \frac{3}{2}$. Then the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$ is:

If $x$ and $y$ are two unit vectors and the angle between them is $\phi$,then $\frac{1}{2} |x - y| = $

If $\bar{a}, \bar{b}$ and $\bar{c}$ are vectors such that $|\bar{a}| = |\frac{\bar{b}}{2}| = |\frac{\bar{c}}{3}| = 1$; $\bar{b}$ and $\bar{c}$ are perpendicular; and the projections of $\bar{b}$ and $\bar{c}$ on $\bar{a}$ are equal,then $|\bar{a} - \bar{b} + \bar{c}| = $

$3 \hat{i}-2 \hat{j}-\hat{k}, -2 \hat{i}-\hat{j}+3 \hat{k}$ and $-\hat{i}+3 \hat{j}-2 \hat{k}$ are the position vectors of the vertices $A, B$ and $C$ of a $\triangle ABC$ respectively. If $H$ is its orthocenter,then $\overrightarrow{HA}+\overrightarrow{HB}+\overrightarrow{HC} = $