Finding the Position Vector of a Point

What a position vector is

The position vector of a point $P(x, y, z)$ is the vector from the origin $O$ to $P$ :

\vec{p} = \overrightarrow{OP} = x\hat{i} + y\hat{j} + z\hat{k}, \qquad |\vec{p}| = \sqrt{x^2 + y^2 + z^2}

Its components are just the point's coordinates, so moving between "the point $(2, 3, 4)$ " and "the vector $2\hat{i} + 3\hat{j} + 4\hat{k}$ " costs nothing. Once every point has a position vector, the vector joining two of them is a subtraction:

\overrightarrow{PQ} = \vec{q} - \vec{p}

That's "head minus tail": the position vector of where the arrow ends, minus where it starts.

The section formula

A point $R$ dividing the segment from $P$ to $Q$ in the ratio $m:n$ has position vector:

\text{Internally: } \dfrac{m\vec{q}+n\vec{p}}{m+n}, \qquad \text{Externally: } \dfrac{m\vec{q}-n\vec{p}}{m-n}

Notice the cross-over: $m$ , the part of the ratio next to $P$ , multiplies $\vec{q}$ , and $n$ multiplies $\vec{p}$ . Internally $R$ lies between $P$ and $Q$ ; externally it lies on the line through them but outside the segment.

Worked example

Find the position vector of $R$ , which divides the segment joining $P$ (position vector $\hat{i}+2\hat{j}-\hat{k}$ ) and $Q$ (position vector $-\hat{i}+\hat{j}+\hat{k}$ ) in the ratio 2:1, both internally and externally.

Solution, internally ( $m=2, n=1$ ):

R = \dfrac{2(-\hat{i}+\hat{j}+\hat{k}) + 1(\hat{i}+2\hat{j}-\hat{k})}{3} = \dfrac{-\hat{i}+4\hat{j}+\hat{k}}{3}

Externally:

R = \dfrac{2(-\hat{i}+\hat{j}+\hat{k}) - 1(\hat{i}+2\hat{j}-\hat{k})}{1} = -3\hat{i}+3\hat{k}

Worked example: a midpoint

Find the position vector of the midpoint of the segment joining $P(2,3,4)$ and $Q(4,1,-2)$ .

Solution: A midpoint is just the 1:1 case of the section formula: average each coordinate:

\left(\dfrac{2+4}{2}, \dfrac{3+1}{2}, \dfrac{4-2}{2}\right) = (3,2,1) = 3\hat{i}+2\hat{j}+\hat{k}

Worked example: the centroid of a triangle

The centroid of a triangle with vertices $\vec{a}$ , $\vec{b}$ , $\vec{c}$ is their average, $\dfrac{\vec{a} + \vec{b} + \vec{c}}{3}$ . Find the centroid of the triangle with vertices $A(1, 2, 3)$ , $B(3, 0, 1)$ and $C(-1, 1, -1)$ .

Solution:

\dfrac{(1 + 3 - 1)\hat{i} + (2 + 0 + 1)\hat{j} + (3 + 1 - 1)\hat{k}}{3} = \hat{i} + \hat{j} + \hat{k}

Worked example: showing three points are collinear

Show that $A(2, -1, 3)$ , $B(4, 3, 1)$ and $C(3, 1, 2)$ lie on one line.

Solution: Find two vectors from $A$ using head minus tail:

\overrightarrow{AB} = 2\hat{i} + 4\hat{j} - 2\hat{k}, \qquad \overrightarrow{AC} = \hat{i} + 2\hat{j} - \hat{k}

$\overrightarrow{AB} = 2\,\overrightarrow{AC}$ , so the two vectors are parallel and share the point $A$ , which puts all three points on one line. In fact $C$ is the midpoint of $AB$ : averaging $A$ and $B$ gives $(3, 1, 2)$ .

Common mistakes

Pairing $m$ with $\vec{p}$ . In $\dfrac{m\vec{q}+n\vec{p}}{m+n}$ , $m$ goes with the far endpoint $Q$ . Swapping them gives the point dividing the segment in $n:m$ instead.
Using $+$ in the external formula. Externally both the numerator and the denominator change to minus: $\dfrac{m\vec{q}-n\vec{p}}{m-n}$ .
Writing $\overrightarrow{PQ} = \vec{p} - \vec{q}$ . That's $\overrightarrow{QP}$ , the same arrow pointing the other way. It's always head minus tail.
Treating a position vector as the point itself in a length question. $|\vec{p}|$ is the distance from the origin. The distance between $P$ and $Q$ is $|\vec{q} - \vec{p}|$ .

Where this comes up in exams

The 2025-26 CBSE Class XII syllabus lists the position vector of a point and the position vector of a point dividing a line segment in a given ratio. The JEE Main 2026 three-dimensional geometry unit lists the section formula.

Checked against CBSE Mathematics curriculum, Classes XI-XII (2025-26) and JEE (Main) 2026 syllabus (NTA). Syllabi can change from year to year, so confirm with the latest official notification for your exam.

Fit it into your week with the free timetable generator.
Mark it off as you revise in the syllabus tracker.
See how long you have left: JEE Main 2027 (Session 1) countdown.