If $O$ is the circumcentre and $O'$ is the orthocentre of the triangle $ABC$,then $\overrightarrow{O'A} + \overrightarrow{O'B} + \overrightarrow{O'C} = $

If $2 \hat{i}-\hat{j}+\hat{k}$ and $\hat{i}-3 \hat{j}-5 \hat{k}$ are the position vectors of the points $A$ and $B$ respectively,$C$ divides $AB$ in the ratio $2:3$ and $M$ is the mid-point of $AB$,then $5(\text{position vector of } C) - 2(\text{position vector of } M) =$

Three vectors $\vec{P}, \vec{Q}$ and $\vec{R}$ are shown in the figure. Let $S$ be any point on the vector $\vec{R}$. The distance between the point $P$ and $S$ is $b|\vec{R}|$. The general relation among vectors $\vec{P}, \vec{Q}$ and $\vec{S}$ is

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors. Then the point of intersection of the line joining the points $\vec{a}+\vec{b}+\vec{c}, \vec{a}-\vec{b}+3 \vec{c}$ and the line joining the points $2 \vec{a}-\vec{b}+\vec{c}, \vec{a}-2 \vec{b}+4 \vec{c}$ is

$2 \hat{i}-3 \hat{j}+\hat{k}$ and $\hat{i}+2 \hat{j}-3 \hat{k}$ are the position vectors of two points $A$ and $B$ respectively and $C$ divides $AB$ in the ratio $3:2$. If $3 \hat{i}-\hat{j}+2 \hat{k}$ is the position vector of a point $D$,then the unit vector in the direction of $\overrightarrow{CD}$ is

Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be two vectors such that $|\overrightarrow{a}| = 2$ and $|\overrightarrow{b}| = 3$. Then the ratio of the projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ to that of $\overrightarrow{b}$ on $\overrightarrow{a}$ is: