Section formula in 3D

3.6.3 · Maths › 3D Geometry

Subtopic: Ek point kaise dhundha jaata hai jo ek line segment ko diye hue ratio mein divide karta hai — 3D space mein.

Core Idea

WHY koi "weighted average" kyun aata hai?

WHAT chahiye: $P = (x, y, z)$ ke coordinates jo line $A B$ par hain aur $A P : P B = m : n$ hai.

HOW — similar triangles / vectors se scratch se derive karo.

Derivation (vector method — sabse clean)

Let position vectors be $a$ (for $A$ ) and $b$ (for $B$ ). Kyunki $P$ segment $A B$ par hai: $A P = \frac{m}{n} P B .$ Yeh step kyun? Ratio $A P : P B = m : n$ ka matlab hai ki $A$ se $P$ tak ka vector, $P$ se $B$ tak ke vector ka $\frac{m}{n}$ guna hai (same direction, internal case).

Position vectors se vectors likhte hain: $p - a = \frac{m}{n} (b - p) .$ Kyun? $A P = p - a$ aur $P B = b - p$ .

$n$ se multiply karo: $n (p - a) = m (b - p) .$ $n p - n a = m b - m p .$ $p$ collect karo: $(m + n) p = m b + n a .$ $p = \frac{m b + n a}{m + n}$

Coordinates read off karte hain:

Is topic ka 80/20

Worked Examples

$m = 2, n = 3$ .
$x = \frac{2 ( 3 ) + 3 ( 1 )}{2 + 3} = \frac{6 + 3}{5} = \frac{9}{5}$ . Kyun? Far weight $m$ , $x_{2}$ ko multiply karta hai.
$y = \frac{2 ( 4 ) + 3 ( - 2 )}{5} = \frac{8 - 6}{5} = \frac{2}{5}$ .
$z = \frac{2 ( - 5 ) + 3 ( 3 )}{5} = \frac{- 10 + 9}{5} = - \frac{1}{5}$ .
$\Rightarrow P = (\frac{9}{5}, \frac{2}{5}, - \frac{1}{5})$ .

$m - n = 3 - 1 = 2$ denominator mein use karo.
$x = \frac{3 ( 4 ) - 1 ( 2 )}{2} = \frac{12 - 2}{2} = 5$ . Subtract kyun? External $\Rightarrow$ $n \to - n$ replace karo.
$y = \frac{3 ( 3 ) - 1 ( 1 )}{2} = \frac{9 - 1}{2} = 4$ .
$z = \frac{3 ( 2 ) - 1 ( 4 )}{2} = \frac{6 - 4}{2} = 1$ .
$\Rightarrow P = (5, 4, 1)$ . Dhyan do ki yeh $B$ se aage hai, jaisa expected tha kyunki $m > n$ .

Forecast: $P$ , $A$ ke kareeb lagta hai, to chhota $m$ vs $n$ expect karo.

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

$z$ se verify karo: $\frac{1 ( - 10 ) + 2 ( - 4 )}{3} = \frac{- 10 - 8}{3} = - 6.$ ✓ Match hua. Ratio confirmed $1 : 2$ — aur indeed $P$ , $A$ ke kareeb hai. Forecast sahi nikla.

Common Mistakes (Steel-man)

Flashcards

Internal section formula for

A, B

in ratio

m : n

P = (\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})

Which endpoint's coordinate gets multiplied by

m

Opposite/far endpoint

B

(yaani

x_{2}

), kyunki weight

m

P

B

ki taraf kheenchta hai.

External division formula change

Replace

n

- n

: denominators

m - n

ban jaate hain aur numerators mein subtract hota hai.

In external division, do

A P

and

P B

point the same or opposite way?

Opposite directions mein (same direction sirf internal division mein hoti hai); yahi reversal

n \to - n

encode karta hai.

Midpoint of

A, B

in 3D

(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2}, \frac{z _{1} + z _{2}}{2})

, yaani ratio

1 : 1

Vector form of section formula

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

How to find ratio in which

P

divides

A B

Ratio

k : 1

set karo, ek coordinate se

k

solve karo, doosre coordinate se verify karo.

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?

Har coordinate independent hai; weighted average alag-alag

x, y, z

par same

m, n

se apply hoti hai.

Recall Feynman: explain to a 12-year-old

Ek stick socho jiske ek end par lal dot ( $A$ ) hai aur doosre par neela dot ( $B$ ). Tum ek aisa nishaan lagana chahte ho ki lal-side ka tukda aur neela-side ka tukda 2 to 3 jaisi ratio mein ho. Us jagah ki position dhundhne ke liye, tum bas dono ends ki ek "mixing" karte ho. Agar nishaan neele ke paas hai, to neele ke numbers zyada count honge. Tum left–right numbers, front–back numbers, aur up–down numbers ko alag-alag mix karte ho — teen chhote mixings, ek har direction ke liye. Bas yahi poora secret hai.

Connections

Distance formula in 3D — verify karo ki dhundha hua point actually $A B$ par hai.
Position vectors — vector derivation $p = \frac{m b + n a}{m + n}$ .
Centroid of a triangle in 3D — section formula ko ratio $2 : 1$ ke saath ek median par apply karo.
Direction ratios and direction cosines — collinearity check same line se follow hoti hai.
Section formula in 2D — yeh parent idea hai, ek dimension kam ke saath.
Equation of a line in 3D — internal/external division line ko parametrize karta hai.