The position vector of point $C$ relative to $B$ is $(\hat{i} + \hat{j})$ and the position vector of $B$ relative to $A$ is $(\hat{i} - \hat{j})$. The position vector of $C$ relative to $A$ is:

Let $\overrightarrow{OA} = \hat{i} - 3\hat{j} + \hat{k}$, $\overrightarrow{OB} = \hat{i} + 3\hat{j} - 2\hat{k}$, and $\overrightarrow{OC} = 4\hat{i} + 3\hat{j} + 5\hat{k}$ be the position vectors of three points $A$, $B$, and $C$. Let $P$ be the point which divides $AB$ in the ratio $2:1$. If $l, m, n$ are the direction cosines of the vector $\overrightarrow{PC}$, then $l + 3m + 2n =$

Given that $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \times \vec{b} = \vec{0}$. What can you conclude about the vectors $\vec{a}$ and $\vec{b}$?

The triad $(x, y, z)$ of real numbers such that $(3 \hat{i}-\hat{j}+2 \hat{k})=(2 \hat{i}+3 \hat{j}-\hat{k}) x+(\hat{i}-2 \hat{j}+2 \hat{k}) y+(-2 \hat{i}+\hat{j}-2 \hat{k}) z$ is

The vector having magnitude of $2 \sqrt{29}$ units in the direction of vector $\vec{a} = 4 \hat{i} + 3 \hat{j} - 2 \hat{k}$ is . . . . . . .

If the position vectors of two points $A$ and $B$ are $\vec{a} + 3\vec{b}$ and $\vec{a} - 2\vec{b}$ respectively,find the position vector of the point that divides $AB$ in the ratio $2:5$.