Section Formula - Vector Algebra

Last Updated : 12 Jun, 2026

The section formula is used to determine the position vector of a point that divides the line segment joining two given points in a specified ratio. In vector algebra, it provides a direct method for finding the coordinates or position vector of the dividing point without measuring distances geometrically.

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Let the position vectors of points P and Q be \overrightarrow{OP}= \overrightarrow{a} and \overrightarrow{OP}= \overrightarrow{b}

Suppose a point R divides the line segment PQ in the ratio m:n in two ways as follows:

1. Internally

When R divides PQ internally. If R divides \overrightarrow{PQ} such that \frac{\overrightarrow{PR}}{\overrightarrow{RQ}} = \frac{m}{n}

where m and n are positive values, we specify that the point R divides \overrightarrow{PQ} internally in the ratio of m : n. 

1

Now from triangles ORQ and OPR, we have

\overrightarrow{PR} = \overrightarrow{OR} - \overrightarrow{OP} = \overrightarrow{r} - \overrightarrow{a}

\overrightarrow{RQ} = \overrightarrow{OQ} - \overrightarrow{OR} = \overrightarrow{b} - \overrightarrow{r}

Hence, we can conclude that,

m (\overrightarrow{b} - \overrightarrow{r})    = n (\overrightarrow{r} - \overrightarrow{a})

On simplification, we get

\overrightarrow{r} = \frac{m\overrightarrow{b}+n\overrightarrow{a}}{m+n}

Midpoint Formula : When the point divides the line segment equally, then m = n

So, we get \overrightarrow{r} = \frac{\overrightarrow{b}+\overrightarrow{a}}{2}

which is called the midpoint formula.

2. Externally

When R divides PQ externally. If R divides \overrightarrow{PQ}  such that \frac{\overrightarrow{PR}}{\overrightarrow{QR}} = \frac{m}{n}

where m and n are positive values, we say that the point R divides \overrightarrow{PQ}  externally in the ratio of the m : n, provided m ≠ n.

2

Now from triangles ORQ and OPR, we have

\overrightarrow{PR} = \overrightarrow{OR} - \overrightarrow{OP} = \overrightarrow{r} - \overrightarrow{a}

\overrightarrow{RQ} = \overrightarrow{OR} - \overrightarrow{OQ} = \overrightarrow{r} - \overrightarrow{b}

Hence, we can conclude that,

m (\overrightarrow{r} - \overrightarrow{b}) = n (\overrightarrow{r} - \overrightarrow{a})

m\overrightarrow{r} - n\overrightarrow{r} = m\overrightarrow{b} - n\overrightarrow{a}

(m-n) \overrightarrow{r} = m\overrightarrow{b} - n\overrightarrow{a}

On simplification, we get

\overrightarrow{r} = \frac{m\overrightarrow{b}-n\overrightarrow{a}}{m-n}

Solved Examples

Problem 1: Find the position vectors of the points which divide the join of the points 2\overrightarrow{a}-3\overrightarrow{b}  and 3\overrightarrow{a}-2\overrightarrow{b}  internally and externally in the ratio 2 : 3.

Let A and B be the given points with the position vectors 2\overrightarrow{a}-3\overrightarrow{b} and  3\overrightarrow{a}-2\overrightarrow{b} respectively.

Let P divide the \overrightarrow{AB}   in the ratio 2 : 3 internally

m = 2 and n = 3

Using internally section formula,

3

Position vector of P =  \frac{m(2\overrightarrow{a}-3\overrightarrow{b})+n(3\overrightarrow{a}-2\overrightarrow{b})}{m+n}

Position vector of P =  \frac{3(2\overrightarrow{a}-3\overrightarrow{b})+2(3\overrightarrow{a}-2\overrightarrow{b})}{2+3}         

Position vector of P =  \frac{6\overrightarrow{a}-9\overrightarrow{b}+6\overrightarrow{a}-4\overrightarrow{b}}{5}

Position vector of P =  \frac{12\overrightarrow{a}}{5}-\frac{13\overrightarrow{b}}{5}

Now, Let P divide the \overrightarrow{AB}    in the ratio 2 : 3 externally

m = 2 and n = 3

Using externally section formula,

4

Position vector of P =  \frac{m(2\overrightarrow{a}-3\overrightarrow{b})-n(3\overrightarrow{a}-2\overrightarrow{b})}{m-n}

Position vector of P =  \frac{3(2\overrightarrow{a}-3\overrightarrow{b})-2(3\overrightarrow{a}-2\overrightarrow{b})}{3-2}

Position vector of P =  \frac{6\overrightarrow{a}-9\overrightarrow{b}-6\overrightarrow{a}+4\overrightarrow{b}}{1}    

Position vector of P = -5\overrightarrow{b}

Problem 2: If \overrightarrow{a} and \overrightarrow{b} are position vectors of points A and B respectively, then find the position vector of points of trisection of AB.

Let P and Q be points of trisection. Then, AP = PQ = QB = k (constant variable)

PB = PQ + QB = k + k = 2k

\frac{AP}{PB} = \frac{k}{2k} = \frac{1}{2}

P divides AB in the ratio 1 : 2

5

Using internally section formula, where m=1 and n=2

Position vector of P =  \frac{m(\overrightarrow{b})+n(\overrightarrow{a})}{m+n}

Position vector of P =  \frac{1(\overrightarrow{b})+2(\overrightarrow{a})}{1+2}

Position vector of P =  \frac{\overrightarrow{b}+2\overrightarrow{a}}{3}

Now, we can clearly see that Q is the mid-point of PB.

Apply mid-point section formula we have,

Position vector of Q =  \frac{\frac{\overrightarrow{b}+2\overrightarrow{a}}{3}+\overrightarrow{b}}{2}

Position vector of Q =  \frac{4\overrightarrow{b}+2\overrightarrow{a}}{6}    

Position vector of Q =  \frac{\overrightarrow{a}+2\overrightarrow{b}}{3}

Practice Problems

  1. Find the midpoint of points having position vectors 2\hat{i}+3\hat{j}\ \text{and}\ 6\hat{i}+\hat{j}
  2. Find the position vector of a point dividing the join of a and b internally in the ratio 3:2.
  3. Find the point dividing the segment joining 2\overrightarrow{a}-\overrightarrow{b}\ \text{and} \ \overrightarrow{a}+3\overrightarrow{b} externally in the ratio 2:1.
  4. Determine the trisection points of the segment joining position vectors \overrightarrow{a}\ \text{and}\ \overrightarrow{b}
  5. Show that the midpoint of the line segment joining points with position vectors \overrightarrow{a} and -\overrightarrow{a} is the origin.
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