Section formula in 3D

Subtopic: How to find a point that divides a line segment in a given ratio — in 3D space.

Core Idea

WHY does a "weighted average" appear?

WHAT we want: coordinates of $P=(x,y,z)$ on line $AB$ with $AP:PB = m:n$ .

HOW — derive from scratch using similar triangles / vectors.

Derivation (vector method — cleanest)

Let position vectors be $\vec{a}$ (for $A$ ) and $\vec{b}$ (for $B$ ). Since $P$ is on segment $AB$ : $\vec{AP} = \frac{m}{n}\,\vec{PB}.$ Why this step? The ratio $AP:PB=m:n$ means the vector from $A$ to $P$ is $\frac{m}{n}$ times the vector from $P$ to $B$ (same direction, internal case).

Write vectors with position vectors: $\vec{p}-\vec{a} = \frac{m}{n}(\vec{b}-\vec{p}).$ Why? $\vec{AP}=\vec{p}-\vec{a}$ and $\vec{PB}=\vec{b}-\vec{p}$ .

Multiply by $n$ : $n(\vec{p}-\vec{a}) = m(\vec{b}-\vec{p}).$ $n\vec{p}-n\vec{a} = m\vec{b}-m\vec{p}.$ Collect $\vec{p}$ : $(m+n)\vec{p} = m\vec{b}+n\vec{a}.$ $\boxed{\;\vec{p} = \frac{m\vec{b}+n\vec{a}}{m+n}\;}$

Reading off coordinates:

The 80/20 of this topic

Worked Examples

$m=2,\ n=3$ .
$x=\dfrac{2(3)+3(1)}{2+3}=\dfrac{6+3}{5}=\dfrac{9}{5}$ . Why? far weight $m$ multiplies $x_2$ .
$y=\dfrac{2(4)+3(-2)}{5}=\dfrac{8-6}{5}=\dfrac{2}{5}$ .
$z=\dfrac{2(-5)+3(3)}{5}=\dfrac{-10+9}{5}=-\dfrac{1}{5}$ .
$\Rightarrow P=\left(\dfrac{9}{5},\dfrac{2}{5},-\dfrac{1}{5}\right)$ .

Use $m-n=3-1=2$ in denominator.
$x=\dfrac{3(4)-1(2)}{2}=\dfrac{12-2}{2}=5$ . Why subtract? external $\Rightarrow$ replace $n\to -n$ .
$y=\dfrac{3(3)-1(1)}{2}=\dfrac{9-1}{2}=4$ .
$z=\dfrac{3(2)-1(4)}{2}=\dfrac{6-4}{2}=1$ .
$\Rightarrow P=(5,4,1)$ . Note it lies beyond $B$ , as expected since $m>n$ .

Forecast: $P$ looks close to $A$ , so expect small $m$ vs $n$ .

Let ratio $= k:1$ . Use $x$ -coordinate: $5=\frac{k(9)+1(3)}{k+1}\Rightarrow 5k+5=9k+3\Rightarrow 2=4k\Rightarrow k=\tfrac12.$ So ratio $=\tfrac12:1 = 1:2$ .

Verify with $z$ : $\dfrac{1(-10)+2(-4)}{3}=\dfrac{-10-8}{3}=-6.$ ✓ Matches. Ratio confirmed $1:2$ — and indeed $P$ is nearer $A$ . Forecast correct.

Common Mistakes (Steel-man)

Flashcards

Internal section formula for

A,B

in ratio

m:n

P=\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n},\frac{mz_2+nz_1}{m+n}\right)

Which endpoint's coordinate gets multiplied by

m

The opposite/far endpoint

B

(i.e.

x_2

), because weight

m

pulls

P

toward

B

External division formula change

Replace

n

-n

: denominators become

m-n

and numerators subtract.

In external division, do

\vec{AP}

and

\vec{PB}

point the same or opposite way?

Opposite directions (they point the same way only in internal division); that reversal is what

n\to -n

encodes.

Midpoint of

A,B

in 3D

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

, i.e. ratio

1:1

Vector form of section formula

\vec p=\frac{m\vec b+n\vec a}{m+n}

How to find ratio in which

P

divides

AB

Set ratio

k:1

, use one coordinate to solve

k

, verify with another coordinate.

What does

m=n

give in external division?

Denominator

m-n=0

→ point at infinity (line through

A,B

direction), undefined point.

Why is 3D section just 1D done thrice?

Each coordinate is independent; the weighted average applies separately to

x,y,z

with same

m,n

Recall Feynman: explain to a 12-year-old

Picture a stick with a red dot at one end ( $A$ ) and blue at the other ( $B$ ). You want to mark a spot so the red-side piece and blue-side piece are in a ratio like 2 to 3. To find that spot's position, you just take a "mixing" of the two ends. If the spot is closer to blue, blue's numbers count more. You mix the left–right numbers, the front–back numbers, and the up–down numbers separately — three little mixings, one for each direction. That's the whole secret.

Connections

Distance formula in 3D — verify a found point actually lies on $AB$ .
Position vectors — the vector derivation $\vec p=\frac{m\vec b+n\vec a}{m+n}$ .
Centroid of a triangle in 3D — apply section formula with ratio $2:1$ on a median.
Direction ratios and direction cosines — collinearity check follows from same line.
Section formula in 2D — this is the parent idea, one dimension fewer.
Equation of a line in 3D — internal/external division parametrizes the line.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, section formula ka idea bilkul simple hai: ek line segment ke do points $A$ aur $B$ hain, aur tumhe wo point $P$ chahiye jo is segment ko $m:n$ ratio me baant de. Iska matlab sirf itna hai ki $P$ ki position do endpoints ka weighted average hai. Jis taraf zyada weight, $P$ us taraf zyada khinch jaata hai — jaise rassi pe agar tum $B$ ke paas khade ho to $B$ ka "vote" zyada count hota hai.

Sabse important baat: 3D bhi 2D/1D jaisa hi hai, bas teen baar karna padta hai — ek baar $x$ ke liye, ek baar $y$ , ek baar $z$ , same $m$ aur $n$ use karke. Formula yaad rakhne ka trick: ratio $AP:PB = m:n$ me, pehla number $m$ hamesha door wale point $B$ ke coordinate ( $x_2$ ) se multiply hota hai. Bahut students ulta kar dete hain ( $m$ ko $A$ ke saath laga dete hain) — yahi sabse common galti hai, isse bacho.

External division me bas $n$ ki jagah $-n$ likh do, denominator $m-n$ ban jaata hai, aur point segment ke bahar aa jaata hai. Yaad rakho: internal division me $\vec{AP}$ aur $\vec{PB}$ ek hi direction me point karte hain, lekin external division me ye opposite directions me hote hain — yahi reversal $n\to -n$ wali sign change me chhupa hai. Agar kisi question me ratio puchha ho, to ratio ko $k:1$ maan ke ek coordinate se $k$ nikaal lo, phir doosre coordinate se verify kar lo — yeh forecast-then-verify trick exam me galti pakad leta hai. Midpoint to bas $1:1$ ka special case hai, simple average.

Core Idea

WHY does a "weighted average" appear?

Derivation (vector method — cleanest)

The 80/20 of this topic

Worked Examples

Common Mistakes (Steel-man)

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — 3D Geometry

Connections