3D Geometry 📡

Lines aur planes teen dimensions mein — direction, distance, aur angle calculate karo!

🏗️ Architect ki Problem

Ek architect ek building design kar raha hai. Usse check karna hai ki do beams (ek floor se ek corner se doosre corner tak) ek doosre se kitni door hain aur kya woh intersect karte hain ya parallel hain ya skew hain (na parallel, na intersecting — space mein kahin aur hain). Yeh real-world 3D geometry problem hai! Class 12 mein hum exactly aisi problems solve karte hain — lines in 3D, angle between them, shortest distance.

1. Direction Cosines aur Direction Ratios

Direction Cosines (l, m, n): actual cos values l = cos α, m = cos β, n = cos γ l² + m² + n² = 1 Direction Ratios (a, b, c): proportional to DCs l/a = m/b = n/c = 1/√(a²+b²+c²) DC from two points P(x₁,y₁,z₁) to Q(x₂,y₂,z₂): l = (x₂-x₁)/PQ, m = (y₂-y₁)/PQ, n = (z₂-z₁)/PQ

2. Equation of a Line in 3D

/* Vector Form */ ⃗r = ⃗a + λ⃗b (point ⃗a se guzarti hai, direction ⃗b hai) /* Cartesian/Symmetric Form */ (x-x₁)/a = (y-y₁)/b = (z-z₁)/c = λ (point (x₁,y₁,z₁), direction ratios a,b,c) /* Two-point form: line through P(x₁,y₁,z₁) and Q(x₂,y₂,z₂) */ (x-x₁)/(x₂-x₁) = (y-y₁)/(y₂-y₁) = (z-z₁)/(z₂-z₁)

3. Angle Between Two Lines

Lines with direction vectors ⃗b₁ and ⃗b₂: cos θ = |⃗b₁·⃗b₂| / (|⃗b₁||⃗b₂|) Lines with direction ratios (a₁,b₁,c₁) and (a₂,b₂,c₂): cos θ = |a₁a₂+b₁b₂+c₁c₂| / √(a₁²+b₁²+c₁²)·√(a₂²+b₂²+c₂²) Perpendicular: a₁a₂+b₁b₂+c₁c₂ = 0 Parallel: a₁/a₂ = b₁/b₂ = c₁/c₂

4. Skew Lines aur Shortest Distance

Skew lines: Woh lines jo na parallel hain, na intersect karti hain — 3D mein alag planes mein hain.

Shortest distance between skew lines = length of common perpendicular

Lines: ⃗r = ⃗a₁ + λ⃗b₁ and ⃗r = ⃗a₂ + μ⃗b₂ Shortest Distance = |(⃗a₂-⃗a₁)·(⃗b₁×⃗b₂)| / |⃗b₁×⃗b₂| /* Cartesian form: */ SD = | x₂-x₁ y₂-y₁ z₂-z₁ | | a₁ b₁ c₁ | | a₂ b₂ c₂ | ÷ |⃗b₁×⃗b₂|

✅ Find SD between: ⃗r=(î+2ĵ+k̂)+λ(2î+ĵ+2k̂) and ⃗r=(2î-ĵ-k̂)+μ(3î-5ĵ-k̂)

⃗a₁=(1,2,1), ⃗a₂=(2,-1,-1), ⃗b₁=(2,1,2), ⃗b₂=(3,-5,-1)
⃗a₂-⃗a₁=(1,-3,-2)
⃗b₁×⃗b₂=î(1×(-1)-2×(-5))-ĵ(2×(-1)-2×3)+k̂(2×(-5)-1×3) = î(-1+10)-ĵ(-2-6)+k̂(-10-3) = (9,8,-13)
|⃗b₁×⃗b₂|=√(81+64+169)=√314
Numerator: (1,−3,−2)·(9,8,−13) = 9−24+26 = 11

SD = 11/√314 units

5. Equation of a Plane

/* General form */ ax + by + cz = d (a,b,c = normal direction) /* Vector form: ⃗r·n̂ = d */ ⃗r·(aî+bĵ+ck̂) = d /* Three-point form: pass through A,B,C */ Use determinant method /* Intercept form: */ x/a + y/b + z/c = 1 Distance from point (x₁,y₁,z₁) to plane ax+by+cz+d=0: d = |ax₁+by₁+cz₁+d| / √(a²+b²+c²)

6. Practice Questions

Q1. Find angle between lines: (x-1)/2=(y+1)/3=(z-4)/6 and (x+1)/1=(y-2)/4=(z-3)/(-7)

Solution

⃗b₁=(2,3,6), ⃗b₂=(1,4,-7). ⃗b₁·⃗b₂=2+12-42=-28. |⃗b₁|=7, |⃗b₂|=√66. cosθ=-28/(7√66)=-4/√66. θ=cos⁻¹(-4/√66)

Q2. Find distance of point (3,-2,1) from plane 2x-y+2z+3=0.

Solution

d=|2(3)+(-1)(-2)+2(1)+3|/√(4+1+4)=|6+2+2+3|/3=13/3 units

Q3. Direction cosines of line joining A(2,1,3) and B(4,3,5).

Solution

AB=(2,2,2), |AB|=2√3. DCs=(2/2√3, 2/2√3, 2/2√3)=(1/√3, 1/√3, 1/√3)