Learning Objectives
Write a line using a point and a nonzero direction vector.
Write a plane using a point and a nonzero normal vector.
Find a line-plane intersection and the distance from a point to a plane.
Prerequisites: vector arithmetic, dot products, and cross products. A line needs a direction along it; a plane is conveniently described by a direction perpendicular to it.
1. Parameterize a Line
A line through P₀ = (x₀, y₀, z₀) with direction v = ⟨a, b, c⟩ has vector equation r(t) = ⟨x₀, y₀, z₀⟩ + t⟨a, b, c⟩, for real t and v ≠ 0. Equivalently, x = x₀ + at, y = y₀ + bt, and z = z₀ + ct.
Worked Example: A Line through Two Points
Find the line through A = (1, 0, 2) and B = (3, −1, 5). Its direction is B − A = ⟨2, −1, 3⟩.
A parameterization is r(t) = ⟨1, 0, 2⟩ + t⟨2, −1, 3⟩.
At t = 0 the point is A; at t = 1 it is B. Restricting 0 ≤ t ≤ 1 gives the segment, while all real t give the full line.
When a direction component is zero, the corresponding coordinate stays constant. Avoid dividing by that component to form a symmetric equation. A parametric equation handles zero components naturally.
2. A Plane Is Perpendicular to Its Normal
If a plane passes through P₀ = (x₀, y₀, z₀) and has nonzero normal n = ⟨a, b, c⟩, then each displacement within the plane is perpendicular to n. Its equation is a(x − x₀) + b(y − y₀) + c(z − z₀) = 0.
Worked Example: A Plane through Three Points
Use A = (0, 0, 1), B = (1, 0, 1), and C = (0, 2, 1). The in-plane vectors are AB = ⟨1, 0, 0⟩ and AC = ⟨0, 2, 0⟩.
Their cross product is ⟨0, 0, 2⟩, a normal vector. Using point A gives 2(z − 1) = 0, or z = 1.
All three original points satisfy z = 1. Three collinear points would not determine a unique plane because the cross product would be zero.
3. Substitute a Line into a Plane
Worked Example: An Intersection
Intersect r(t) = ⟨1, 0, 2⟩ + t⟨2, −1, 3⟩ with the plane x + y + z = 7.
Substitute the coordinates: (1 + 2t) + (−t) + (2 + 3t) = 7. Thus 3 + 4t = 7 and t = 1.
The intersection is (3, −1, 5). Check: 3 − 1 + 5 = 7.
If substitution produces a contradiction such as 3 = 7, the line is parallel to the plane and disjoint from it. If it produces an identity, every point on the line lies in the plane. Otherwise, solving the resulting linear equation gives one intersection.
4. Distance to a Plane
For a plane ax + by + cz + d = 0 and a point P = (x₁, y₁, z₁), the perpendicular distance is |ax₁ + by₁ + cz₁ + d|/√(a² + b² + c²). The denominator divides by the normal's length, so scaling a plane equation does not change the distance.
Example: Distance to a Horizontal Plane
The distance from (2, −1, 5) to z − 1 = 0 is |5 − 1|/1 = 4. This agrees with the vertical separation between z = 5 and z = 1.
Common Mistakes
A line direction lies along the line; a plane normal is perpendicular to the plane.
Use the same parameter value in all three coordinates when finding an intersection.
Keep the absolute value in the distance formula. Distance cannot be negative.
5. Practice
1. Parameterize the line through (0, 2, −1) in direction ⟨3, 0, 2⟩.
2. Find a plane through (1, 2, 3) with normal ⟨2, −1, 4⟩.
3. Find where r(t) = ⟨t, 2t, 3t⟩ meets z = 6, and the distance from the origin to z = 6.
Worked Solutions
1. x = 3t, y = 2, z = −1 + 2t, with t real. The zero y component makes y constant.
2. 2(x − 1) − (y − 2) + 4(z − 3) = 0, or 2x − y + 4z = 12. Substituting the given point gives 12 on the left.
3. 3t = 6 gives t = 2 and intersection (2, 4, 6). The perpendicular distance from the origin to the plane is 6, not the length of the slanted line segment to that intersection.
6. Summary
- A point and direction specify a line; a point and normal specify a plane.
- Substitute a line's coordinates into a plane equation to test intersection.
- Point-to-plane distance measures separation in the normal direction.