Equations of planes

A flat surface in space

A plane is a flat surface that extends forever. In 3D, a plane can be described in two main ways.

Vector equation of a plane

A point and two direction vectors (non-parallel, lying in the plane):

r = r₀ + s u + t v

Example: Through A(1, 2, 0) with u = [1, 0, 2] and v = [0, 1, −1]:

[x, y, z] = [1, 2, 0] + s[1, 0, 2] + t[0, 1, −1]

Each pair (s, t) gives a point on the plane. This is the vector equation of a plane.

Scalar equation of a plane

A normal vector n = [A, B, C] is perpendicular to every vector in the plane.

Scalar equation of a plane: Ax + By + Cz + D = 0

The coefficients A, B, C are the normal vector’s components.

From a point and a normal

Example: Through P(2, −1, 3) with normal n = [4, 1, −2]

• 4x + y − 2z + D = 0
• Substitute P: 8 − 1 − 6 + D = 0 → D = −1
• 4x + y − 2z − 1 = 0

From the vector equation to the scalar equation

Find the normal with the cross product of the direction vectors.

From the earlier example: n = [1, 0, 2] × [0, 1, −1]

• = [0(−1) − 2(1), 2(0) − 1(−1), 1(1) − 0(0)] = [−2, 1, 1]
• −2x + y + z + D = 0; through (1, 2, 0): −2 + 2 + 0 + D = 0 → D = 0
• −2x + y + z = 0 (or 2x − y − z = 0)

From three points

1. Make two direction vectors from the points.
2. Cross them to get n.
3. Substitute a point to find D.

Example: A(1, 0, 0), B(0, 2, 0), C(0, 0, 3)

• AB⃗ = [−1, 2, 0], AC⃗ = [−1, 0, 3]
• n = AB⃗ × AC⃗ = [6, 3, 2]
• 6x + 3y + 2z + D = 0 → through A: 6 + D = 0 → D = −6
• 6x + 3y + 2z = 6

Intercepts

For 6x + 3y + 2z = 6: x-intercept 1, y-intercept 2, z-intercept 3. Plotting the three intercepts helps sketch the plane.

Special planes

Parallel and perpendicular planes

• Parallel planes have parallel normal vectors.
• Perpendicular planes have normal vectors with dot product 0.

Is a point on the plane?

Substitute. (1, 1, 1) in 6x + 3y + 2z = 6? 11 ≠ 6 → no.

Worked examples

Example 1: Normal vector of 3x − y + 5z = 7? [3, −1, 5].

Example 2: Plane through the origin with normal [1, 1, 1]? x + y + z = 0.

Example 3: Are 2x + y − z = 3 and 4x + 2y − 2z = 1 parallel? Normals [2, 1, −1] and [4, 2, −2] → yes.

Common mistakes

• Using parallel direction vectors (they don’t define a plane).
• Sign errors in the cross product.
• Forgetting to find D.

Try it yourself

1. Plane through (1, 2, 3) with normal [2, −1, 4].
2. Normal to x − 2y + 3z = 0?
3. Plane through A(0, 0, 0), B(1, 0, 0), C(0, 1, 0)?
4. Are x + y + z = 1 and x − y = 0 perpendicular?

Answers: 1) 2x − y + 4z − 12 = 0 2) [1, −2, 3] 3) z = 0 4) yes ([1, 1, 1] · [1, −1, 0] = 0)

Quick recap

• A plane has a vector equation (point + two direction vectors) and a scalar equation Ax + By + Cz + D = 0.
• [A, B, C] is the normal vector.
• Find normals with the cross product.