Section formula

Two points, a ratio, and one vector formula — that is all it takes to pin down a third point sitting anywhere on the line joining them. Given AA and BB by their position vectors, the section formula hands you the position vector of the point RR that splits ABAB in a stated ratio m:nm:n, whether RR sits between AA and BB (internal division) or beyond one of them on the line produced (external division). It replaces a slower coordinate-geometry setup — similar triangles, or solving simultaneous equations for a dividing point — with a single weighted average of a⃗\vec a and b⃗\vec b. The formula sits right after vector addition and scalar multiplication in the ISC Class 12 syllabus, because it is really nothing more than those two operations combined: scale each endpoint's position vector by the "opposite" part of the ratio, add, and normalise by m+nm+n (or m−nm-n externally). Master it and midpoints, dividing points, and even working backward to an unknown endpoint all become one short calculation.

The core insight

The position vector of the point dividing segment ABAB in the ratio m:nm:n is a weighted average of a⃗\vec a and b⃗\vec b: internally r⃗=mb⃗+na⃗m+n\vec r = \dfrac{m\vec b+n\vec a}{m+n}, externally r⃗=mb⃗−na⃗m−n\vec r = \dfrac{m\vec b-n\vec a}{m-n}.

The core idea

Let AA and BB be points with position vectors a⃗\vec a and b⃗\vec b, measured from a fixed origin OO. Suppose RR divides the segment ABAB internally in the ratio m:nm:n, meaning AR:RB=m:nAR:RB = m:n with RR lying between AA and BB. The position vector of RR is

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

where: a⃗,b⃗\vec a, \vec b are the position vectors of the two endpoints, m,nm, n are the (positive) parts of the ratio AR:RBAR:RB, and r⃗\vec r is the position vector of the dividing point RR. Notice the pairing: nn multiplies a⃗\vec a and mm multiplies b⃗\vec b — each endpoint's vector is weighted by the ratio-part measured at the other endpoint, not its own. If instead RR lies on the line ABAB produced, outside the segment, so that the division is external in the ratio m:nm:n, the same bookkeeping with a subtraction gives

r⃗=mb⃗−na⃗m−n,(m≠n).\vec r = \frac{m\vec b - n\vec a}{m-n}, \qquad (m \ne n).

A very common special case is the midpoint of ABAB: setting m=n=1m=n=1 in the internal formula collapses it to

r⃗=a⃗+b⃗2.\vec r = \frac{\vec a + \vec b}{2}.

Once r⃗\vec r is written in component form as xi^+yj^+zk^x\hat i + y\hat j + z\hat k, the coordinates of RR can be read off directly — the i^,j^,k^\hat i, \hat j, \hat k coefficients of a position vector are the point's x,y,zx, y, z coordinates, so no extra conversion step is needed.

Figure
Segment AB with point R dividing it internally in the ratio m to n, and the position vectors a, b, r drawn from a fixed origin O
R divides AB internally in the ratio m to n O A (a) B (b) R (r) m n n : m written along AR : RB reads m : n r = (m b + n a) / (m + n)
Derivation ADeriving the internal section formula

The goal is to show, using only vector addition and scalar multiplication (no coordinate geometry, no calculus), that the point RR dividing segment ABAB internally so that AR:RB=m:nAR:RB = m:n must have position vector r⃗=(mb⃗+na⃗)/(m+n)\vec r = (m\vec b + n\vec a)/(m+n), starting purely from the geometric statement that RR lies on segment ABAB with that ratio.

Step 1. Express the condition AR:RB=m:nAR:RB = m:n as a vector equation. Since RR lies on segment ABAB between AA and BB, the vectors AR⃗\vec{AR} and RB⃗\vec{RB} point the same way along the line, and their magnitudes are in ratio m:nm:n. So n AR⃗=m RB⃗n\,\vec{AR} = m\,\vec{RB} (cross-multiplying the ratio, both sides now equal vectors of the same length and direction).

Step 2. Write AR⃗\vec{AR} and RB⃗\vec{RB} in terms of position vectors. Since AR⃗=r⃗−a⃗\vec{AR} = \vec r - \vec a and RB⃗=b⃗−r⃗\vec{RB} = \vec b - \vec r, substitute into Step 1's equation: n(r⃗−a⃗)=m(b⃗−r⃗)n(\vec r - \vec a) = m(\vec b - \vec r).

Step 3. Expand and collect all r⃗\vec r terms on one side. nr⃗−na⃗=mb⃗−mr⃗n\vec r - n\vec a = m\vec b - m\vec r, so nr⃗+mr⃗=mb⃗+na⃗n\vec r + m\vec r = m\vec b + n\vec a, giving (m+n)r⃗=mb⃗+na⃗(m+n)\vec r = m\vec b + n\vec a.

Step 4. Divide through by (m+n)(m+n), which is nonzero since m,nm, n are positive ratio parts. This isolates r⃗=mb⃗+na⃗m+n\vec r = \dfrac{m\vec b + n\vec a}{m+n} — exactly the internal section formula. The external formula follows the identical steps with AR⃗\vec{AR} and RB⃗\vec{RB} pointing in opposite directions (so the ratio equation carries a minus sign instead), producing (m−n)r⃗=mb⃗−na⃗(m-n)\vec r = m\vec b - n\vec a.

Worked example

Find the position vector of the point RR that divides the segment joining A(2,−1,3)A(2, -1, 3) and B(4,3,−1)B(4, 3, -1) internally in the ratio 3:23:2. Also state the coordinates of RR.

Step 1. Write the position vectors of the endpoints. a⃗=2i^−j^+3k^\vec a = 2\hat i - \hat j + 3\hat k and b⃗=4i^+3j^−k^\vec b = 4\hat i + 3\hat j - \hat k.

Step 2. Apply the internal section formula with m=3m=3, n=2n=2 (since RR divides ABAB in ratio 3:23:2): r⃗=3b⃗+2a⃗3+2=3(4i^+3j^−k^)+2(2i^−j^+3k^)5\vec r = \dfrac{3\vec b + 2\vec a}{3+2} = \dfrac{3(4\hat i + 3\hat j - \hat k) + 2(2\hat i - \hat j + 3\hat k)}{5}.

Step 3. Simplify the numerator. 3b⃗=12i^+9j^−3k^3\vec b = 12\hat i + 9\hat j - 3\hat k and 2a⃗=4i^−2j^+6k^2\vec a = 4\hat i - 2\hat j + 6\hat k, so 3b⃗+2a⃗=16i^+7j^+3k^3\vec b + 2\vec a = 16\hat i + 7\hat j + 3\hat k.

Step 4. Divide by 5 to get r⃗=165i^+75j^+35k^\vec r = \dfrac{16}{5}\hat i + \dfrac{7}{5}\hat j + \dfrac{3}{5}\hat k. So RR has coordinates (165,75,35)\left(\dfrac{16}{5}, \dfrac{7}{5}, \dfrac{3}{5}\right).

Competitive stretch

Working backward: if instead P(4,1,2)P(4,1,2) is known to divide ABAB internally in ratio 2:12:1 with A(2,−2,2)A(2,-2,2), the same formula can be solved for the unknown b⃗\vec b. Writing p⃗=2b⃗+a⃗3\vec p = \dfrac{2\vec b + \vec a}{3} and rearranging gives b⃗=3p⃗−a⃗2=(12i^+3j^+6k^)−(2i^−2j^+2k^)2=5i^+52j^+2k^\vec b = \dfrac{3\vec p - \vec a}{2} = \dfrac{(12\hat i+3\hat j+6\hat k) - (2\hat i-2\hat j+2\hat k)}{2} = 5\hat i + \tfrac52\hat j + 2\hat k, so B=(5,52,2)B = \left(5, \dfrac52, 2\right) — the section formula runs equally well forwards (find RR) or backwards (find a missing endpoint).

Common traps

  • Swapping mm and nn: for AR:RB=m:nAR:RB = m:n, the numerator is mb⃗+na⃗m\vec b + n\vec a — each ratio part pairs with the other point's vector — so writing ma⃗+nb⃗m\vec a + n\vec b by reflex is a very common slip.
  • Forgetting to divide by m+nm+n (internal) or m−nm-n (external) after combining the vectors.
  • Using the internal-division formula when the problem actually describes external division, or vice versa — check whether RR lies between AA and BB or on the line produced.
  • Confusing the position vectors a⃗,b⃗\vec a, \vec b (measured from the origin) with the displacement vector AB⃗=b⃗−a⃗\vec{AB} = \vec b - \vec a; the section formula needs the former, not the latter.

Ask the tutor: want to see how the section formula is used to find the ratio in which a known point divides a segment, rather than the dividing point itself?

Why does the internal formula use m b plus n a instead of m a plus n b?

It comes straight from the ratio AR:RB=m:nAR:RB = m:n. In the derivation, n(r⃗−a⃗)=m(b⃗−r⃗)n(\vec r-\vec a) = m(\vec b-\vec r) rearranges to (m+n)r⃗=mb⃗+na⃗(m+n)\vec r = m\vec b + n\vec a, so mm (the part nearer AA) ends up multiplying b⃗\vec b, and nn (the part nearer BB) ends up multiplying a⃗\vec a — each ratio-part pairs with the other endpoint's vector. The safest way to remember it without re-deriving: the ratio-part written first (mm, next to AA in "m:nm:n") multiplies the other point's vector, b⃗\vec b.

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

Read the geometry, not just the ratio. If RR is described as lying on segment ABAB, between AA and BB, or the problem simply says "divides ABAB in the ratio m:nm:n" with no further qualifier, use internal division. If it says RR lies on ABAB produced, or outside the segment, or the ratio is negative in context, use external division with the minus signs. When in doubt, sketch the three points roughly to see where RR should fall.

Can the section formula be used to prove things like "the diagonals of a parallelogram bisect each other"?

That kind of general geometric proof is outside this syllabus unit — here the section formula is used for direct computation only: finding a dividing point, a midpoint, or an unknown endpoint or ratio from given position vectors. Save proof-style applications for a full vector-geometry or coordinate-geometry course if you meet them later; for this unit, expect numerical position vectors in, numerical position vectors out.

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Where this leads