Section formula

A GPS app plotting a rest stop exactly $40\%$ of the way along a road trip, or a physics problem finding the balance point between two masses — both are really asking the same vector question: given two points, where does the point sitting in a stated ratio between them land? The section formula answers this in one line, turning a geometric ratio into position-vector arithmetic. It is built from nothing more than the triangle law of vector addition and scalar multiplication, yet it is one of the most exam-frequent results in the Class 12 Vector Algebra chapter — the midpoint formula and the triangle centroid both fall out of it as special cases. Get the weight-pairing right once, and internal division, external division, midpoints and centroids all become the same three-line calculation.

The core insight

The point dividing $AB$ internally in ratio $m:n$ has position vector $\vec r = \dfrac{m\vec b+n\vec a}{m+n}$ , obtained from the triangle law by walking from $A$ a fraction $\frac{m}{m+n}$ of the way toward $B$ .

The core idea

Fix an origin $O$ , and let points $A$ and $B$ have position vectors $\vec a$ and $\vec b$ . Let $R$ be the point on segment $AB$ that divides it internally in the ratio $m:n$ , meaning $AR:RB = m:n$ . Its position vector is

\vec r = \frac{m\vec b + n\vec a}{m+n},

where: $\vec a,\vec b$ are the position vectors of the two endpoints, $m,n$ are the two positive parts of the ratio (with $m$ paired to $\vec b$ , the far point from $A$ ), and $\vec r$ is the position vector of the dividing point. A useful way to hold this formula in memory is the "cross-multiply" pairing: the coefficient sitting next to $\vec a$ is $n$ (the part of the ratio nearer $B$ ), and the coefficient next to $\vec b$ is $m$ (the part nearer $A$ ). As $R$ slides toward $B$ , $m$ grows relative to $n$ , and the formula automatically leans the answer toward $\vec b$ — a fast sanity check against a swapped ratio. Three situations fall directly out of this one idea.

External division. If $R$ lies on line $AB$ extended, outside the segment, dividing it externally in ratio $m:n$ , the same bookkeeping with a subtraction instead of an addition gives

\vec r = \frac{m\vec b - n\vec a}{m-n}, \qquad (m\neq n).

Midpoint. Setting $m=n=1$ collapses the section formula to the midpoint of $AB$ :

\vec m = \frac{\vec a+\vec b}{2}.

Centroid of a triangle. For a triangle with vertices $A,B,C$ (position vectors $\vec a,\vec b,\vec c$ ), the centroid $G$ divides each median in ratio $2:1$ from the vertex, which simplifies to the clean symmetric form

\vec g = \frac{\vec a+\vec b+\vec c}{3}.

Notice how each of these three results is really the same formula wearing a different hat: the midpoint sets the ratio to $1:1$ , and the centroid result comes from applying the section formula twice — once to find the midpoint of a side, and once more along the median in ratio $2:1$ from the opposite vertex. There is genuinely only one idea to learn here, not four.

Figure

Segment AB with point R dividing it internally in ratio m to n, showing the weighted pull toward B

The main derivation

Derivation ASection formula for internal division

The goal is to derive the position vector of the point $R$ that divides segment $AB$ internally in ratio $m:n$ using only the triangle law of vector addition and scalar multiplication of a vector, starting from the geometric statement $AR:RB=m:n$ and ending at the closed-form $\vec r = (m\vec b+n\vec a)/(m+n)$ that every later special case (midpoint, centroid) reduces to.

Step 1. Translate the ratio into a vector statement. Since $R$ lies on segment $AB$ with $AR:RB=m:n$ , the length $AR$ is the fraction $\frac{m}{m+n}$ of the whole segment $AB$ . Because $\overrightarrow{AR}$ and $\overrightarrow{AB}$ point the same way along the line, this fraction carries over to the vectors themselves: $\overrightarrow{AR} = \dfrac{m}{m+n}\,\overrightarrow{AB}$ .

Step 2. Write $\overrightarrow{AB}$ in terms of the position vectors. By the triangle law, $\overrightarrow{AB} = \vec b - \vec a$ (walking from $A$ to $O$ then $O$ to $B$ reverses the first leg). Substituting into Step 1 gives $\overrightarrow{AR} = \dfrac{m}{m+n}(\vec b-\vec a)$ .

Step 3. Locate $R$ from the origin. Again by the triangle law, $\vec r = \vec a + \overrightarrow{AR}$ , so $\vec r = \vec a + \dfrac{m}{m+n}(\vec b-\vec a)$ .

Step 4. Combine over a common denominator. $\vec r = \dfrac{(m+n)\vec a + m\vec b - m\vec a}{m+n} = \dfrac{m\vec b + n\vec a}{m+n}$ , since the $m\vec a$ terms cancel, leaving only $n\vec a$ alongside $m\vec b$ . This is the section formula.

Worked example

Find the position vector of the point $R$ that divides the segment joining $A(\vec a = 2\hat i-3\hat j+4\hat k)$ and $B(\vec b = -\hat i+2\hat j-\hat k)$ internally in the ratio $3:2$ .

Step 1. Identify $m$ and $n$ . The ratio $AR:RB=3:2$ gives $m=3$ , $n=2$ .

Step 2. Apply the formula. $\vec r = \dfrac{m\vec b+n\vec a}{m+n} = \dfrac{3(-\hat i+2\hat j-\hat k)+2(2\hat i-3\hat j+4\hat k)}{5}$ .

Step 3. Expand each term. $3(-\hat i+2\hat j-\hat k) = -3\hat i+6\hat j-3\hat k$ , and $2(2\hat i-3\hat j+4\hat k) = 4\hat i-6\hat j+8\hat k$ . Adding component-wise: $(-3+4)\hat i+(6-6)\hat j+(-3+8)\hat k = \hat i+0\hat j+5\hat k$ .

Step 4. Divide by $m+n=5$ . $\vec r = \dfrac{\hat i+5\hat k}{5} = \dfrac{1}{5}\hat i+\hat k$ .

Answer: $R$ has position vector $\dfrac{1}{5}\hat i+\hat k$ .

Competitive stretch

The same machinery verifies a centroid two ways. For a triangle with $\vec a=\hat i+2\hat j+3\hat k$ , $\vec b=3\hat i-\hat j+2\hat k$ , $\vec c=-\hat i+4\hat j-\hat k$ , the direct centroid formula gives $\vec g = \dfrac{\vec a+\vec b+\vec c}{3} = \hat i+\dfrac{5}{3}\hat j+\dfrac{4}{3}\hat k$ . Checking via the median from $A$ : the midpoint of $BC$ is $\vec d=\dfrac{\vec b+\vec c}{2}=\hat i+\dfrac32\hat j+\dfrac12\hat k$ , and applying the section formula on $AD$ in ratio $2:1$ from $A$ gives $\vec g=\dfrac{2\vec d+\vec a}{3}=\dfrac{3\hat i+5\hat j+4\hat k}{3}$ , matching exactly — the two routes to a centroid always agree.

Common traps

Swapping which weight goes with which point. The ratio $AR:RB=m:n$ pairs $m$ with $\vec b$ and $n$ with $\vec a$ — writing $\dfrac{n\vec b+m\vec a}{m+n}$ by accident flips the point toward the wrong end.
Using the internal-division formula for an external-division problem. External division needs a minus sign in both numerator and denominator: $\vec r=\dfrac{m\vec b-n\vec a}{m-n}$ , not just a sign flip in the numerator alone.
Treating the midpoint formula as unrelated. It is simply $m=n=1$ in the same formula; students who memorize it separately sometimes apply the wrong ratio pairing on a genuine section-formula question.
Deriving the centroid via a median and mis-ordering the $2:1$ split. The direct $\vec g=\frac{\vec a+\vec b+\vec c}{3}$ is the safest route; use the median only to verify.

Ask the tutor: stuck on whether a problem needs internal or external division, or want the section-formula derivation walked through step by step again? Ask the AI tutor.

Why does the ratio m:n pair m with b instead of a?

Because $m$ measures the part of the segment closer to $B$ : in $AR:RB=m:n$ , $m$ is the length $AR$ 's "share" and it is exactly the fraction of $\overrightarrow{AB}$ you travel from $A$ . When you finish the derivation and simplify, that travel fraction ends up multiplying $\vec b$ , while the leftover fraction, tied to $n$ , multiplies $\vec a$ . The pairing feels backward until you re-derive it once by hand; after that, the "far point gets $m$ " rule sticks.

How do I know if a problem wants internal or external division?

Read whether the dividing point lies between $A$ and $B$ (internal) or outside the segment, on the extended line (external). Problems usually say "divides externally" explicitly, or give a ratio with $m>n$ combined with a point clearly beyond $B$ . If unsure, compute both interpretations and check which position vector actually lies on the correct side using the given coordinates — a quick visual or numeric check catches most mistakes.

Is there a quicker way to remember the centroid formula without deriving it every time?

Yes — treat $\vec g=\frac{\vec a+\vec b+\vec c}{3}$ as a standalone fact worth memorizing directly, the way you would memorize the midpoint formula. Reserve the median-based derivation (section formula in ratio $2:1$ ) for when a question explicitly asks you to prove the centroid divides each median in that ratio, or when you want a fast cross-check on an already-computed answer. In practice, most board questions simply hand you three vertices and ask for the centroid directly, so the one-line formula is usually all the marks require; the median route is worth keeping only because it occasionally appears as its own separate proof question.