Section Formula in Vector Form — Divide in Ratio 2:3

Question

Points $A$ and $B$ have position vectors $\vec{a}$ and $\vec{b}$ respectively. Find the position vector of point $P$ that divides $AB$ internally in the ratio $2:3$.

Solution — Step by Step

Why This Works

The position vector of any point is just its “address” from the origin $O$. When $P$ divides $AB$ in ratio $2:3$, it means $AP = 2k$ and $PB = 3k$ for some scalar $k$ — so $P$ sits two-fifths of the way from $A$ to $B$.

We can derive the formula from scratch: $\vec{OP} = \vec{OA} + \vec{AP}$. Since $\vec{AP} = \frac{m}{m+n}\vec{AB}$, we get $\vec{OP} = \vec{a} + \frac{2}{5}(\vec{b} - \vec{a})$, which simplifies to $\frac{3\vec{a} + 2\vec{b}}{5}$. Same answer, different path.

This formula works for all position vectors regardless of the coordinate system. In CBSE 12 and JEE Main, this exact setup (with variables $\vec{a}$ and $\vec{b}$, no coordinates) is the standard form — memorise it cold.

Alternative Method

Using the parametric approach:

$\vec{OP} = \vec{OA} + \vec{AP}$

Since $P$ divides $AB$ in $2:3$, point $P$ is at $\frac{2}{5}$ of the way from $A$ to $B$:

Common Mistake