Visual walkthrough — Section formula in 3D

We will not use a single symbol before it is drawn. Let's start with nothing but two points.

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Step 1 — Two points and a straight rope

WHAT. We place two points in space and call them $A$ and $B$. A point is just a location; in 3D we pin it down with three numbers — how far right ($x$), how far forward ($y$), how far up ($z$).

WHY. Everything about "dividing a segment" lives on the straight line joining $A$ and $B$. So the first thing we need is that straight line — the rope pulled tight between two pegs.

PICTURE. Look at the figure: the amber dot is $A$, the cyan dot is $B$, and the white line is the taut rope. The little box on $A$ shows its three coordinates $(x_1,y_1,z_1)$.

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Step 2 — Mark a spot and name the two pieces

WHAT. We stop somewhere on the rope and call that stopping-spot $P$. It cuts the rope into two pieces: the piece from $A$ to $P$, and the piece from $P$ to $B$.

WHY. "Dividing a segment in a ratio" is a statement about the lengths of these two pieces. We must name them before we can compare them.

PICTURE. The white bracket labels the near piece $AP$; the cyan bracket labels the far piece $PB$. $P$ is the amber square sitting on the rope.

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Step 3 — Trade lengths for arrows (position vectors)

WHAT. Instead of dragging around three coordinates at once, we bundle each point's three numbers into a single arrow from the origin $O$ out to that point. That arrow is the point's position vector.

WHY THIS TOOL. Lengths alone ($AP$, $PB$) don't remember direction. But dividing a segment needs both "how far" and "which way". An arrow carries both at once, so an arrow equation can handle all three coordinates in one line — this is the payoff of Position vectors. We reach for vectors precisely because they let us avoid writing $x$, $y$, $z$ three separate times during the algebra.

PICTURE. Three arrows leave the origin: $\vec{a}$ to $A$ (amber), $\vec{b}$ to $B$ (cyan), $\vec{p}$ to $P$ (white). The rope $AB$ is the same line as before, now floating out in space away from $O$.

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Step 4 — The one honest equation about $P$

WHAT. The arrow along the rope from $A$ to $P$ is $\vec{p}-\vec{a}$. The arrow from $P$ to $B$ is $\vec{b}-\vec{p}$. Because $P$ sits between $A$ and $B$, these two arrows point the same way, and the first is $\frac{m}{n}$ times as long as the second: $\vec{p}-\vec{a}=\frac{m}{n}(\vec{b}-\vec{p}).$

WHY. "Arrow from tip to tip = (head position) − (tail position)" is the one rule of vector subtraction. And $\frac{AP}{PB}=\frac{m}{n}$ with same direction is exactly the statement $\overrightarrow{AP}=\frac{m}{n}\overrightarrow{PB}$.

PICTURE. Term by term:

$\underset{\text{arrow }A\to P,\text{length }AP}{\underbrace{\vec{p}-\vec{a}}}=\underset{\text{stretch factor}}{\underbrace{\frac{m}{n}}}\underset{\text{arrow }P\to B,\text{length }PB}{\underbrace{(\vec{b}-\vec{p})}}$

The figure shows both arrows nose-to-tail along the rope, both cyan-tinted to signal "same direction", with the stretch factor $\frac{m}{n}$ labelled between them.

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Step 5 — Untangle the algebra to isolate $\vec{p}$

WHAT. We solve the arrow equation for $\vec{p}$, one clean move at a time.

WHY. We want a formula — "$\vec{p}$ equals stuff we already know ($\vec{a},\vec{b},m,n$)". So we sweep every $\vec{p}$ to one side.

Multiply both sides by $n$ (clears the fraction): $n(\vec{p}-\vec{a})=m(\vec{b}-\vec{p}).$ Open the brackets: $\underset{P\text{ terms}}{\underbrace{n\vec{p}}}-n\vec{a}=m\vec{b}-\underset{P\text{ terms}}{\underbrace{m\vec{p}}}.$ Gather the two $\vec{p}$ terms on the left, the known arrows on the right: $n\vec{p}+m\vec{p}=m\vec{b}+n\vec{a}.$ Factor out $\vec{p}$: $(m+n)\vec{p}=m\vec{b}+n\vec{a}.$ Divide by the number $(m+n)$: $\boxed{\vec{p}=\frac{m\vec{b}+n\vec{a}}{m+n}}$

PICTURE. The figure tracks the migration: $\vec{p}$ terms (white) sliding from both sides into one heap, the known arrows (amber $\vec{a}$, cyan $\vec{b}$) settling on the other. The final boxed arrow lands on the rope, its position governed by the weights.

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Step 6 — Why the weights are crossed (the "far weight" picture)

WHAT. In $\vec{p}=\frac{m\vec{b}+n\vec{a}}{m+n}$, the weight $m$ sits on $\vec{b}$ (the point $B$), not on $\vec{a}$.

WHY. Trace it back to Step 4: a large $m$ means $AP$ is long, so $P$ has travelled a long way from $A$ — it now leans toward $B$. A point leaning toward $B$ must weight $B$'s coordinates more. So $m$ belongs to $B$. The near-looking pairing "$m$ with $A$" is the classic trap.

PICTURE. Two panels. Left: $m:n=1:4$ — tiny weight on $B$, so $P$ hugs $A$. Right: $m:n=4:1$ — heavy weight on $B$, so $P$ hugs $B$. The dot visibly leans toward whichever side carries more weight.

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Step 7 — The midpoint: equal weights

WHAT. Set $m=n$ (say both $=1$). Then $P$ is the exact middle.

WHY. Equal weights mean neither end pulls harder, so the average is the plain halfway point. This is the sanity anchor for the whole formula.

$M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$

PICTURE. The dot rests dead-centre; both bracketed pieces $AM$ and $MB$ are drawn equal.

Recall

Why does $m=n$ give the midpoint? ::: Equal weights ⇒ the weighted average is the plain average of coordinates ⇒ halfway.

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Step 8 — When $P$ slips outside the rope (external division)

WHAT. Now let $P$ sit on the line but beyond $B$ (or behind $A$). The two arrows $\overrightarrow{AP}$ and $\overrightarrow{PB}$ no longer point the same way — they point in opposite directions.

WHY. In Step 4 we assumed same direction. Once $P$ is outside, $\overrightarrow{PB}$ reverses. A reversal is captured by flipping the sign: write the ratio as $m:(-n)$, i.e. replace $n\to -n$ everywhere.

$P=\left(\frac{m{x}_2-n{x}_1}{m-n},\frac{m{y}_2-n{y}_1}{m-n},\frac{m{z}_2-n{z}_1}{m-n}\right)$

The numerators subtract, and the denominator becomes $m-n$.

PICTURE. The rope $AB$ is drawn, then extended past $B$ with a dashed line to the external $P$. The arrow $\overrightarrow{PB}$ (cyan) now points back toward $B$, opposite to $\overrightarrow{AP}$ (amber) — the opposing arrowheads are the whole story.

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The one-picture summary

WHAT. One figure holds the entire story: same rope, three positions of $P$ — hugging $A$ (small $m$), dead-centre ($m=n$), hugging $B$ (large $m$), and one external $P$ beyond $B$ — each labelled with the weighted-average formula that produces it.

$\vec{p}=\frac{m\vec{b}+n\vec{a}}{m+n}\stackrel{n\to -n}{\to }\vec{p}=\frac{m\vec{b}-n\vec{a}}{m-n}$

Recall Feynman retelling of the whole walkthrough

Stretch a rope between a red peg $A$ and a blue peg $B$. Pick a spot $P$ on it. The rope splits into a near piece $AP$ and a far piece $PB$; their length-comparison $m:n$ describes where $P$ sits. To pin down $P$'s actual location, replace each peg by an arrow from the corner of the room (the origin) — arrows remember both distance and direction, so one arrow-equation does the work of three coordinate equations. That equation just says "the arrow from $A$ to $P$ is $\frac{m}{n}$ of the arrow from $P$ to $B$." Tidy the algebra and out pops a weighted average: $\vec{p}=(m\vec{b}+n\vec{a})/(m+n)$. The surprise is the crossing — the weight $m$ rides on $B$, because a long near-piece means $P$ has walked toward $B$. Equal weights land you dead centre (the midpoint). And if you let $P$ slip off the end of the rope, one arrow flips direction; that flip is bookkept by turning $n$ into $-n$, which makes the pluses into minuses and the bottom into $m-n$. If $m=n$ out there, the bottom is zero and $P$ has escaped to infinity. That's the entire section formula, seen rather than memorised.

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Connections

• Position vectors — the arrows $\vec{a},\vec{b},\vec{p}$ that made Step 4 possible.
• Section formula in 2D — the same walkthrough with one arrow-component fewer.
• Distance formula in 3D — check a found $P$ really lies on $AB$ (its two pieces sum to $AB$).
• Centroid of a triangle in 3D — apply this with ratio $2:1$ along a median.
• Equation of a line in 3D — sliding $P$ (letting $m:n$ vary) is the parametrised line.
• Direction ratios and direction cosines — same-direction test underlying Steps 4 and 8.

Concept Map