3D Geometry

Distance between P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) in space. Direct extension of the 2D formula using Pythagoras twice.

Point dividing PQ internally in ratio m:n. For external division, replace n with −n. Midpoint is the special case m = n = 1.

If a line makes angles α, β, γ with the positive x, y, z axes respectively, the direction cosines are l = cos α, m = cos β, n = cos γ, and they always satisfy l²+m²+n²=1.

If (a, b, c) are direction ratios of a line, the direction cosines are obtained by dividing by √(a²+b²+c²). Direction ratios are proportional to direction cosines but not normalized.

Acute angle θ between two lines with direction cosines (l₁, m₁, n₁) and (l₂, m₂, n₂). The absolute value ensures the acute angle is taken.

Angle between two lines with direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂). Derived by normalizing DRs to DCs first.

Two lines with direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂) are perpendicular iff their dot product is zero. Equivalently l₁l₂+m₁m₂+n₁n₂ = 0.

Two lines are parallel iff their direction ratios are proportional. The angle between them is 0°.

Line through point with position vector a, parallel to vector b. λ ∈ ℝ is the parameter. Every point on the line corresponds to a unique λ.

Line through (x₁, y₁, z₁) with direction ratios (a, b, c). Each ratio equals the parameter λ. If any direction ratio is zero (say a = 0), the corresponding equation is x = x₁.

Line through P(x₁, y₁, z₁) and Q(x₂, y₂, z₂). Direction ratios are (x₂−x₁, y₂−y₁, z₂−z₁).

Distance from point P (position vector p) to the line r = a + λb. The cross product |b × (a−p)| gives the area of the parallelogram with sides b and (a−p); dividing by |b| gives the perpendicular height.

For skew lines r = a₁+λb₁ and r = a₂+μb₂, the shortest distance is the projection of (a₂−a₁) onto the common perpendicular direction b₁×b₂.

Two lines r = (x₁,y₁,z₁)+λ(a₁,b₁,c₁) and r = (x₂,y₂,z₂)+μ(a₂,b₂,c₂) are coplanar iff this determinant vanishes. Equivalently: (a₂−a₁)·(b₁×b₂) = 0.

Every first-degree equation in x, y, z represents a plane in 3D. The normal to the plane has direction ratios (a, b, c). The vector form is r·n = −d where n = (a, b, c).

Plane making intercepts p, q, r on the x, y, z axes respectively. Passes through (p, 0, 0), (0, q, 0), (0, 0, r).

Equation of the plane passing through three non-collinear points P₁(x₁,y₁,z₁), P₂, P₃. The determinant expresses the coplanarity of (x,y,z) with the three given points.

Plane at perpendicular distance p from the origin, with unit normal (l, m, n) where l²+m²+n²=1 and p > 0. To convert ax+by+cz+d=0: divide by ±√(a²+b²+c²) choosing sign so p > 0.

Acute dihedral angle between planes a₁x+b₁y+c₁z+d₁=0 and a₂x+b₂y+c₂z+d₂=0. The angle between planes equals the angle between their normals (or its supplement — take the acute one).

Angle θ between line with DRs (a₁,b₁,c₁) and plane ax+by+cz+d=0. Note sin (not cos) because θ is measured from the plane, not from the normal. Complementary to the angle between the line and the normal.

Perpendicular distance from point P(x₁, y₁, z₁) to plane ax+by+cz+d=0. The formula is the 3D analogue of the 2D point-to-line distance.

The foot of the perpendicular from P(x₁,y₁,z₁) to plane ax+by+cz+d=0. The perpendicular from P has direction (a,b,c) (the normal); substitute the parametric point into the plane equation to find the parameter.

Image (reflection) of P(x₁,y₁,z₁) in plane ax+by+cz+d=0. The image P′ is such that the foot of perpendicular is the midpoint of PP′. The formula follows by doubling the foot-of-perpendicular displacement.

Every plane through the line of intersection of planes P₁=0 and P₂=0, for varying λ. One additional condition (e.g., passes through a given point, is perpendicular to a given plane) determines λ.