Distance formula in 3D — Hinglish notes

3.6.2 · Maths › 3D Geometry

YEH HAI KYA?

YEH SACH KYUN HAI? (Scratch se Derivation)

Hum ise khud banate hain — formula par blindly trust kabhi mat karo.

Step 1 — Ek box mein daalo. $P$ aur $Q$ ko ek axis-aligned rectangular box (cuboid) ke opposite corners maano. Is box ke teen edges ki lengths hain: $a = ∣ x_{2} - x_{1} ∣, b = ∣ y_{2} - y_{1} ∣, c = ∣ z_{2} - z_{1} ∣$

Yeh step kyun? Kyunki har coordinate mein jo difference hai wahi box ke woh edges hain jo axes ke parallel hain — yeh "distance" ko right angles ki geometry mein convert kar deta hai jo hum pehle se samajhte hain.

Step 2 — Floor par Pythagoras. Sirf box ke bottom face ko dekho. Is rectangle ka diagonal $d$ (jo $x y$ -plane mein hai) satisfy karta hai: $d^{2} = a^{2} + b^{2}$

Yeh step kyun? Donon edges $a$ aur $b$ $9 0^{\circ}$ par milti hain (x-axis ⟂ y-axis), toh floor diagonal ekdum plain 2D Pythagoras hai.

Step 3 — Pythagoras upar ki taraf. Ab body diagonal $P Q$ ek naye right triangle ki hypotenuse hai jiske legs hain floor diagonal $d$ aur vertical edge $c$ . Vertical edge poore floor ke perpendicular hai, isliye: $P Q^{2} = d^{2} + c^{2} = (a^{2} + b^{2}) + c^{2}$

Yeh step kyun? Z-direction poore $x y$ -plane ke perpendicular hai, isliye $d$ ke bhi perpendicular hai jo us plane mein hai — toh phir se Pythagoras cleanly apply hota hai.

Step 4 — Wapas substitute karo. $P Q = a^{2} + b^{2} + c^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$

Note: squaring karne se absolute-value signs khatam ho jaate hain, toh final form mein $∣ \cdot ∣$ ki zaroorat nahi.

USE KAISE KARTE HAIN — worked examples

Common mistakes

Recall Feynman: ek 12-saal ke bachche ko samjhao

Ek shoebox imagine karo. Ek chhoti si chinti uske ek bottom corner mein baithi hai aur ek crumb opposite top corner mein hai. Chinti box ke andar se seedha nahi chal sakti — lekin hum us seedhi diagonal ko measure kar sakte hain. Pehle floor ke across measure karo (box ki length aur width use karke ek Pythagoras). Phir top corner tak tilt karo (us floor-diagonal aur box ki height use karke ek aur Pythagoras). Donon ko stack karo aur ek neat rule milta hai: length, width, aur height ke differences ko square karo, add karo, square root lo. Woh number real straight-line gap hai.

Active Recall

P (x_{1}, y_{1}, z_{1})

aur

Q (x_{2}, y_{2}, z_{2})

ke beech 3D distance formula kya hai?

(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}

Point

(x, y, z)

ki origin se distance?

x^{2} + y^{2} + z^{2}

Distance formula mein subtraction ki order matter kyun nahi karti?

Kyunki har difference square hota hai, aur squaring sign hata deta hai.

3D distance derive karne ke liye Pythagoras kitni baar apply hota hai?

Do baar — ek baar floor diagonal ke liye, ek baar body diagonal tak jaane ke liye.

Distances use karke teen points A, B, C ki collinearity kaise test karte hain?

Check karo ki sabse badi distance baaki donon ka sum hai ya nahi (jaise

A B + B C = A C

(1, - 2, 3)

aur

(4, 2, 15)

ke beech distance?

13

Distance

P Q

kis vector ka magnitude hai?

P Q = (x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1})

Agar

(x, 2, - 1)

(2, 4, - 2)

se distance 3 par hai, toh

x

nikalo.

x = 0

x = 4

Connections

Coordinate axes and octants in 3D — coordinates jo is formula ko feed karte hain.
Section formula in 3D — inhi coordinate differences ko use karta hai.
Vectors and magnitude — distance = $∣ P Q ∣$ .
Distance formula in 2D — special case jahan $z$ drop ho jaata hai.
Direction cosines and direction ratios — inhi differences se bane hain.
Equation of a sphere — saare points ek centre se fixed distance par.