The position vectors of the points $A, B, C$ are $(2\hat{i}+\hat{j}-\hat{k}), (3\hat{i}-2\hat{j}+\hat{k})$ and $(\hat{i}+4\hat{j}-3\hat{k})$ respectively. These points

If $\vec{a}$ and $\vec{b}$ are the position vectors of $A$ and $B$,respectively,find the position vector of a point $C$ in $BA$ produced such that $BC = 1.5 BA$.

Let the vectors $\overrightarrow{AB} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\overrightarrow{AC} = 2\hat{i} + 4\hat{j} + 4\hat{k}$ be two sides of a triangle $ABC$. If $G$ is the centroid of $\triangle ABC$, then $\frac{27}{7}(\overrightarrow{AG})^2 + 5 =$

If the position vectors of the points $A, B, C$ are $i + j$,$i - j$,and $a i + b j + c k$ respectively,then the points $A, B, C$ are collinear if

Let $\bar{a}, \bar{b}$ and $\bar{c}$ be non-coplanar vectors. If $P, Q, R$ and $S$ are four points with position vectors $-\bar{a}+4\bar{b}-3\bar{c}$,$3\bar{a}+2\bar{b}-5\bar{c}$,$-3\bar{a}+8\bar{b}-5\bar{c}$ and $-3\bar{a}+2\bar{b}+\bar{c}$ respectively,then the ordered pair $(x, y)$ of real numbers such that $\overline{PQ} = x \cdot \overline{PR} + y \cdot \overline{PS}$ is

If the points with position vectors $10\hat{i} + 3\hat{j}$,$12\hat{i} - 5\hat{j}$,and $a\hat{i} + 11\hat{j}$ are collinear,then the value of $a$ is: