Section formula in 3D: Worked examples — Maths deep dive

This page is the drill ground for the Section formula in 3D. The parent note built the formula; here we throw every kind of situation at it — internal, external, midpoint, the "find-the-ratio" reverse, a degenerate case that blows up, a real-world word problem, and an exam twist. By the end you should never meet a version of this question you haven't already seen solved.

Before we start, let's re-anchor the two symbols we lean on constantly:

And the two formulas we'll reuse (from the parent):

The scenario matrix

Every question this topic can throw belongs to one of these cells. The worked examples are labelled by cell so you can see the whole map get covered.

Cell	What makes it special	Covered by
C1 Internal, all positive coords	the "default" mixing	Ex 1
C2 Internal, mixed-sign coords	negatives in $A$ or $B$ — sign bookkeeping	Ex 2
C3 Midpoint ( $m = n = 1$ )	the halfway special case	Ex 3
C4 External, $m > n$	$P$ lands beyond $B$	Ex 4
C5 External, $m < n$	$P$ lands behind $A$ (denominator negative)	Ex 5
C6 Degenerate external, $m = n$	denominator $= 0$ → point at infinity	Ex 6
C7 Reverse: find the ratio	given $P$ , solve for $k : 1$ (incl. external sign)	Ex 7
C8 Real-world word problem	dividing a physical path/pipe in a ratio	Ex 8
C9 Exam twist: unknown coordinate	one endpoint coord missing, $P$ given	Ex 9

Let's walk them.

Ex 1 — Cell C1 · Internal, all positive

Forecast: ratio $2 : 1$ means the $A$ -to- $P$ piece is twice the $P$ -to- $B$ piece — so $P$ is two-thirds of the way from $A$ to $B$ , i.e. much closer to $B$ . Guess $P$ near $B$ 's side. Look at the number line in the figure below and predict before reading.

Identify weights. $m = 2$ (touches $A$ -side), $n = 1$ . Why this step? The formula needs $m, n$ ; getting them straight prevents the classic " $m$ goes with $A$ " slip — remember far weight: $m$ multiplies $B$ 's coordinate.
$x$ : $\frac{2 ( 8 ) + 1 ( 2 )}{2 + 1} = \frac{16 + 2}{3} = \frac{18}{3} = 6$ . Why? far weight $m$ multiplies $x_{2}$ .
$y$ : $\frac{2 ( 6 ) + 1 ( 3 )}{3} = \frac{12 + 3}{3} = 5$ .
$z$ : $\frac{2 ( 7 ) + 1 ( 1 )}{3} = \frac{14 + 1}{3} = 5$ . Why separate lines? $x, y, z$ are three independent 1D mixings; never blend them.
Answer: $P = (6, 5, 5)$ .

Verify: $P$ should be $\frac{2}{3}$ of the way from $A$ . Check $x$ : $2 + \frac{2}{3} (8 - 2) = 2 + 4 = 6$ ✓. And $P = (6, 5, 5)$ sits nearer $B (8, 6, 7)$ than $A (2, 3, 1)$ , matching our forecast.

Ex 2 — Cell C2 · Internal with negatives

Forecast: $3 : 2$ is only slightly more than $1 : 1$ , so $P$ should sit a bit past the midpoint, leaning toward $B$ . Watch the minus signs — they trip most people.

Weights: $m = 3, n = 2$ , sum $m + n = 5$ . Why? we'll divide by $5$ each time.
$x$ : $\frac{3 ( 2 ) + 2 ( - 4 )}{5} = \frac{6 - 8}{5} = - \frac{2}{5}$ . Why the minus? $x_{1} = - 4$ carries its sign into the product $n x_{1} = 2 (- 4) = - 8$ ; don't drop it.
$y$ : $\frac{3 ( - 4 ) + 2 ( 2 )}{5} = \frac{- 12 + 4}{5} = - \frac{8}{5}$ .
$z$ : $\frac{3 ( 3 ) + 2 ( - 6 )}{5} = \frac{9 - 12}{5} = - \frac{3}{5}$ .
Answer: $P = (- \frac{2}{5}, - \frac{8}{5}, - \frac{3}{5})$ .

Verify (midpoint sanity): midpoint of $A, B$ is $(- 1, - 1, - \frac{3}{2})$ . Since $m > n$ , $P$ should be nudged from the midpoint toward $B (2, - 4, 3)$ : indeed $x$ moved from $- 1$ up to $- 0.4$ (toward $B$ 's $+ 2$ ), $z$ moved from $- 1.5$ up to $- 0.6$ (toward $B$ 's $+ 3$ ). ✓ Forecast confirmed.

Ex 3 — Cell C3 · Midpoint ( $m = n = 1$ )

Forecast: midpoint = plain average of each coordinate, no weighting. Guess each coord is halfway.

Recognise the special case. $m = n = 1 \Rightarrow$ formula collapses to $(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2}, \frac{z _{1} + z _{2}}{2})$ . Why? with equal weights the $m$ 's cancel and we just average.
$x$ : $\frac{1 + 7}{2} = 4$ .
$y$ : $\frac{- 3 + 3}{2} = 0$ .
$z$ : $\frac{5 + ( - 1 )}{2} = 2$ .
Answer: $M = (4, 0, 2)$ .

Verify: $M$ must be equidistant from $A$ and $B$ . Using the Distance formula in 3D: $A M^{2} = (4 - 1)^{2} + (0 + 3)^{2} + (2 - 5)^{2} = 9 + 9 + 9 = 27$ and $M B^{2} = (7 - 4)^{2} + (3 - 0)^{2} + (- 1 - 2)^{2} = 9 + 9 + 9 = 27$ . Equal ✓.

Ex 4 — Cell C4 · External, $m > n$ ( $P$ beyond $B$ )

Forecast: external + $m > n$ means $P$ shoots past $B$ , on the far side. It is not between the pegs. See the figure.

Switch to external formula: replace $n \to - n$ , so denominator is $m - n = 3 - 1 = 2$ . Why? externally $A P$ and $P B$ point opposite ways; writing $A P : P B = 3 : (- 1)$ encodes that reversal.
$x$ : $\frac{3 ( 4 ) - 1 ( 1 )}{2} = \frac{12 - 1}{2} = \frac{11}{2} = 5.5$ .
$y$ : $\frac{3 ( 5 ) - 1 ( 2 )}{2} = \frac{15 - 2}{2} = \frac{13}{2} = 6.5$ .
$z$ : $\frac{3 ( 6 ) - 1 ( 3 )}{2} = \frac{18 - 3}{2} = \frac{15}{2} = 7.5$ .
Answer: $P = (5.5, 6.5, 7.5)$ .

Verify: every coordinate of $P$ exceeds $B$ 's ( $5.5 > 4$ , $6.5 > 5$ , $7.5 > 6$ ), confirming $P$ lies beyond $B$ . Also $P$ is collinear with $A, B$ : the direction $B - A = (3, 3, 3)$ and $P - A = (4.5, 4.5, 4.5) = 1.5 (3, 3, 3)$ — same direction, so on the line ✓ (see Direction ratios and direction cosines).

Ex 5 — Cell C5 · External, $m < n$ ( $P$ behind $A$ )

Forecast: external with $m < n$ → denominator $m - n$ is negative, and $P$ lands on the $A$ side, behind $A$ (opposite direction to $B$ ). Guess coordinates smaller than $A$ 's.

External formula: $m - n = 1 - 3 = - 2$ . Why keep the negative? a negative denominator is legal — it just flips the mixing to the far side of $A$ ; don't take absolute value.
$x$ : $\frac{1 ( 6 ) - 3 ( 2 )}{- 2} = \frac{6 - 6}{- 2} = \frac{0}{- 2} = 0$ .
$y$ : $\frac{1 ( 4 ) - 3 ( 0 )}{- 2} = \frac{4 - 0}{- 2} = - 2$ .
$z$ : $\frac{1 ( 5 ) - 3 ( 1 )}{- 2} = \frac{5 - 3}{- 2} = \frac{2}{- 2} = - 1$ .
Answer: $P = (0, - 2, - 1)$ .

Verify: $P$ 's coordinates fall below $A (2, 0, 1)$ ( $0 < 2$ , $- 2 < 0$ , $- 1 < 1$ ), so $P$ is on the far side of $A$ away from $B$ ✓. Collinearity: $P - A = (- 2, - 2, - 2) = - 0.5 (4, 4, 4) = - 0.5 (B - A)$ — negative multiple, i.e. opposite direction, exactly "behind $A$ " ✓.

Ex 6 — Cell C6 · Degenerate external, $m = n$ (point at infinity)

Forecast: $m = n$ externally makes the denominator $m - n = 0$ . Dividing by zero — you should expect no finite point (the point "runs off to infinity"). Predict "undefined".

Set up denominator: $m - n = 2 - 2 = 0$ . Why is this the whole story? every coordinate would be $\frac{something}{0}$ .
$x$ : $\frac{2 ( 1 ) - 2 ( 0 )}{0} = \frac{2}{0}$ — undefined. Why? geometrically, asking for a point equally far out on both sides of the segment forces $P$ infinitely far along the line.
Interpretation: external ratio $1 : 1$ describes the direction of the line (a point at infinity), not a real location.
Answer: no finite $P$ exists; the "point" is at infinity along direction $B - A = (1, 1, 1)$ .

Verify: the numerator $2 (1) - 2 (0) = 2 \neq = 0$ while the denominator is $0$ , so the limit as the two weights approach equality diverges — a genuine $\frac{nonzero}{0}$ blow-up, confirming "at infinity" and never $\frac{0}{0}$ (which would be indeterminate). ✓

Ex 7 — Cell C7 · Reverse problem: find the ratio

Forecast: $P = (1, - 2, - 3)$ — its $x = 1$ sits between $A$ 's $3$ and $B$ 's $- 1$ , so $P$ looks between the pegs (internal), roughly in the middle. Guess a ratio near $1 : 1$ .

Assume ratio $k : 1$ . Why $k : 1$ ? any ratio $m : n$ equals $\frac{m}{n} : 1$ , so one unknown $k$ suffices; sign of $k$ will reveal internal (+) vs external (−).
Use the $x$ -equation: $1 = \frac{k ( - 1 ) + 1 ( 3 )}{k + 1}$ . Why $x$ ? any single coordinate determines $k$ ; pick the simplest.
Solve: $1 (k + 1) = - k + 3 \Rightarrow k + 1 = - k + 3 \Rightarrow 2 k = 2 \Rightarrow k = 1$ . So ratio $= 1 : 1$ (the midpoint).
Answer: $P$ divides $A B$ internally in $1 : 1$ — it is the midpoint.

Verify with $y$ and $z$ : midpoint $y = \frac{4 + ( - 8 )}{2} = - 2$ ✓; midpoint $z = \frac{- 1 + ( - 7 )}{2} = - 4$ … wait, that gives $- 4$ , but $P$ 's $z = - 3$ . Check again carefully: using $k = 1$ , $z = \frac{1 ( - 7 ) + 1 ( - 1 )}{2} = \frac{- 8}{2} = - 4 \neq = - 3$ .

To keep the example honest, take the corrected point $P^{'} (1, - 2, - 4)$ (its $z$ fixed to lie on the line). Redo the check: $x$ gives $k = 1$ , $y = - 2$ ✓, $z = - 4$ ✓. Ratio confirmed $1 : 1$ for $P^{'}$ .

Ex 8 — Cell C8 · Real-world word problem

Forecast: $2 : 3$ leans toward $A$ (the $A$ -piece is smaller), so the sensor is less than halfway, nearer $A$ . Guess coordinates below the midpoint.

Model it: the sensor $S$ divides $A B$ internally with $m = 2, n = 3$ , sum $= 5$ . Why internal? the sensor is on the pipe between the junctions.
$x$ : $\frac{2 ( 30 ) + 3 ( 0 )}{5} = \frac{60}{5} = 12$ m.
$y$ : $\frac{2 ( 12 ) + 3 ( 0 )}{5} = \frac{24}{5} = 4.8$ m.
$z$ : $\frac{2 ( - 6 ) + 3 ( 0 )}{5} = \frac{- 12}{5} = - 2.4$ m.
Answer: drill at $S = (12, 4.8, - 2.4)$ metres.

Verify (units + proportion): since $A$ is the origin, $S$ should be exactly $\frac{2}{5} = 0.4$ of the way to $B$ . Check: $0.4 \times (30, 12, - 6) = (12, 4.8, - 2.4)$ ✓. Coordinates are metres throughout (consistent units), and $S$ is nearer $A$ as forecast.

Ex 9 — Cell C9 · Exam twist: unknown endpoint coordinate

Forecast: we're given the ratio and some coordinates; use whichever coordinate is fully known to sanity-check, then solve for the unknowns. Ratio $1 : 2$ → $P$ nearer $A$ .

Weights: $m = 1, n = 2$ , sum $= 3$ . Why? standard internal setup; denominator $3$ .
Confirm with the fully-known $x$ : $\frac{1 ( 7 ) + 2 ( 1 )}{3} = \frac{9}{3} = 3$ . But $P$ 's $x = 4$ ?! This flags an inconsistency — so re-read: perhaps $P$ 's $x$ pins the ratio. Instead, solve for the ratio from $x$ : $4 = \frac{k ( 7 ) + 1}{k + 1} \Rightarrow 4 k + 4 = 7 k + 1 \Rightarrow 3 = 3 k \Rightarrow k = 1$ , ratio $1 : 1$ . Why switch? the given "ratio $1 : 2$ " conflicts with $x$ , so trust the coordinate — the intended ratio is $1 : 1$ (midpoint).
Find $y_{P}$ with ratio $1 : 1$ : $y_{P} = \frac{3 + 9}{2} = 6$ .
Find $z_{A}$ with ratio $1 : 1$ and $P$ 's $z = 2$ : $2 = \frac{z _{A} + 8}{2} \Rightarrow z_{A} + 8 = 4 \Rightarrow z_{A} = - 4$ .
Answer: $y_{P} = 6, z_{A} = - 4$ (with the corrected ratio $1 : 1$ ).

Verify: midpoint of $A (1, 3, - 4)$ and $B (7, 9, 8)$ is $(4, 6, 2)$ — exactly $P$ ✓ across all three coordinates.

Recall check

Recall

Which cell has a negative denominator? ::: External with $m < n$ (Ex 5): $m - n < 0$ , placing $P$ behind $A$ . External $m = n$ gives what? ::: Denominator $0$ → point at infinity, undefined (Ex 6). How to find the ratio $P$ divides $A B$ ? ::: Set $k : 1$ , solve from one coordinate, verify with the other two (Ex 7). If $x, y$ give one ratio but $z$ gives another? ::: $P$ is not on the line at all — the three must agree. Real-world $2 : 5$ -style path split? ::: Same internal formula, keep units consistent (Ex 8).

Connections

Section formula in 3D — the parent formula these examples drill.
Distance formula in 3D — used to verify midpoint (Ex 3) and collinearity checks.
Position vectors — the vector origin of $p = \frac{m b + n a}{m + n}$ .
Direction ratios and direction cosines — the "same/opposite direction" collinearity tests in Ex 4, Ex 5.
Centroid of a triangle in 3D — extend Ex-style ratio $2 : 1$ on medians.
Section formula in 2D — drop $z$ and every example collapses to its 2D twin.
Equation of a line in 3D — internal/external division as a parametrization of the line.