Linear Geometry
3.2 Equation of a Plane
Normal vector equation of a plane, scalar form, distance from point to plane and from origin, angle between planes, line-plane intersection, three-point plane, parametric form.
3.2 Equation of a Plane
Normal vector form: A plane through point A with normal n:
n·(r−a) = 0 → n₁x + n₂y + n₃z = d
where d = n₁a₁ + n₂a₂ + n₃a₃ = n·a
n·(r−a) = 0 → n₁x + n₂y + n₃z = d
where d = n₁a₁ + n₂a₂ + n₃a₃ = n·a
Example: Plane through A(1,2,3) with normal n=(2,−1,4):
2(x−1)−1(y−2)+4(z−3) = 0
→ 2x − y + 4z = 12
2(x−1)−1(y−2)+4(z−3) = 0
→ 2x − y + 4z = 12
3.2.1 Distance from Origin to Plane
d = |D||n| where plane is n·r = D
3.2.2 Distance from Point P to Plane
d = |n·p − D||n| where plane is n·r = D
Example: Distance from P(1,0,2) to plane 2x−y+4z=12:
d = |2(1)−1(0)+4(2)−12|4+1+16 = |2+8−12|21 = 221 ≈ 0.436
d = |2(1)−1(0)+4(2)−12|4+1+16 = |2+8−12|21 = 221 ≈ 0.436
3.2.3 Angle Between Two Planes
cos θ = |n₁·n₂||n₁||n₂|
Two planes are parallel if n₁ = λn₂. They are perpendicular if n₁·n₂ = 0.
3.2.4 Plane Through Three Points
Given A, B, C on the plane: n = AB×AC (normal = cross product of two edge vectors)
Example: Plane through A(1,0,0), B(0,2,0), C(0,0,3):
AB = (−1,2,0), AC = (−1,0,3)
n = AB×AC = (6,3,2)
Equation: 6x + 3y + 2z = 6·1 = 6
AB = (−1,2,0), AC = (−1,0,3)
n = AB×AC = (6,3,2)
Equation: 6x + 3y + 2z = 6·1 = 6
3.2.5 Intersection of Line and Plane
Substitute parametric line r = a+td into plane equation, solve for t, then find point.
3.2.6 Parametric Equation of a Plane
r = a + su + tv
where u and v are two non-parallel vectors in the plane (s,t ∈ ℝ)
where u and v are two non-parallel vectors in the plane (s,t ∈ ℝ)