Equation of a Plane

The equation of a plane through the point (x₀, y₀, z₀) with normal vector n = ⟨a, b, c⟩ is a(x − x₀) + b(y − y₀) + c(z − z₀) = 0, which simplifies to ax + by + cz = d. The normal vector is perpendicular to every direction in the plane. For three points, find n with a cross product.

Before this lesson: Cross Product, Equations of Lines in 3D and Dot Product

What determines a plane

A line in space is set by a point and a direction. A plane is set by a point and a normal vector, a vector that sticks straight out of the plane, perpendicular to it.

Here is why that works. If P₀ is a fixed point in the plane and P is any other point in it, the arrow from P₀ to P lies flat in the plane. So it is perpendicular to n, and their dot product is 0. Writing that dot product out in coordinates gives the equation of the plane.

The coefficients of x, y and z in ax + by + cz = d are the components of a normal vector. So you can read n = ⟨a, b, c⟩ straight from the equation.

Equation of a plane formulas

\mathbf{n} \cdot (\mathbf{r} - \mathbf{r}_0) = 0

a(x - x_0) + b(y - y_0) + c(z - z_0) = 0

ax + by + cz = d, \quad \text{where } d = ax_0 + by_0 + cz_0

D = \dfrac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}} \quad \text{(distance from } (x_1, y_1, z_1) \text{ to the plane)}

• Point-normal form a(x − x₀) + b(y − y₀) + c(z − z₀) = 0 needs one point and a normal vector.
• For three points A, B and C, use n = AB × AC.
• Two planes are parallel when their normal vectors are parallel. The angle between two planes is the angle between their normals.
• The intercepts of ax + by + cz = d are found by setting two of the variables equal to 0.

Equation of a plane examples

Example 1: Point and normal vector

Find the equation of the plane through P(1, −2, 4) with normal vector n = ⟨3, 1, −2⟩.

3(x - 1) + 1(y + 2) - 2(z - 4) = 0

3x - 3 + y + 2 - 2z + 8 = 0

3x + y - 2z + 7 = 0

3x + y - 2z = -7

3 \cdot 1 + (-2) - 2 \cdot 4 = 3 - 2 - 8 = -7

Answer: 3x + y − 2z = −7.

Example 2: Plane through three points

Find the equation of the plane through P(1, 0, 2), Q(3, 1, 0) and R(0, 2, 3).

\overrightarrow{PQ} = \langle 2, 1, -2 \rangle, \quad \overrightarrow{PR} = \langle -1, 2, 1 \rangle

\overrightarrow{PQ} \times \overrightarrow{PR} = \langle 5, 0, 5 \rangle

\mathbf{n} = \langle 1, 0, 1 \rangle

1(x - 1) + 0(y - 0) + 1(z - 2) = 0

x - 1 + z - 2 = 0

x + z = 3

3 + 0 = 3, \quad 0 + 3 = 3

Answer: x + z = 3. The plane contains every value of y, so it is parallel to the y-axis.

Example 3: Intercepts

Find the x-, y- and z-intercepts of the plane 2x + 3y + 6z = 12, and sketch it.

2x = 12

x = 6

3y = 12

y = 4

6z = 12

z = 2

Answer: the intercepts are (6, 0, 0), (0, 4, 0) and (0, 0, 2). Connecting them gives the triangle in the diagram, the part of the plane in the first octant.

Example 4: Where a line meets a plane

Find where the line x = 2 + t, y = 1 − t, z = 2t meets the plane 2x + y + z = 11.

2(2 + t) + (1 - t) + 2t = 11

4 + 2t + 1 - t + 2t = 11

5 + 3t = 11

3t = 6

t = 2

(2 + 2, \; 1 - 2, \; 2 \cdot 2) = (4, -1, 4)

2 \cdot 4 + (-1) + 4 = 11

Answer: the line meets the plane at (4, −1, 4).

Example 5: Camera distance from a roof (word problem)

A sloped roof lies in the plane x + 2y + 2z = 10, measured in meters. How far is a security camera at (4, 6, 9) from the plane of the roof?

1 \cdot 4 + 2 \cdot 6 + 2 \cdot 9 - 10 = 4 + 12 + 18 - 10

= 24

\sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3

D = \dfrac{|24|}{3} = 8

Answer: the camera is 8 m from the plane of the roof, measured along the perpendicular.

Common mistakes

Related methods

Angle between two planes

The angle between two planes is the angle between their normals. For x + z = 3 and y + z = 1, the normals are ⟨1, 0, 1⟩ and ⟨0, 1, 1⟩:

1 \cdot 0 + 0 \cdot 1 + 1 \cdot 1 = 1

\cos\theta = \dfrac{1}{\sqrt{2} \cdot \sqrt{2}} = \dfrac{1}{2}

\theta = \cos^{-1}\left(\dfrac{1}{2}\right) = 60^\circ

Line of intersection of two planes

Two planes that aren’t parallel meet in a line. Its direction is n₁ × n₂, since that vector is perpendicular to both normals. Find one point on both planes (for example, by setting z = 0 and solving), then write the line as in equations of lines in 3D. The distance formula above is a vector projection onto n in disguise.

Practice problems

1. Find the equation of the plane through (2, 1, −3) with normal ⟨4, −1, 5⟩.
2. Give a normal vector to the plane 3x − 7y + z = 4.
3. Find the equation of the plane through A(1, 1, 0), B(2, 0, 1) and C(0, 3, 1).
4. Find the intercepts of 4x + 2y + 5z = 20.
5. Is the point (1, 3, −2) on the plane 2x − y + 4z = −9?
6. Find the plane through (3, −1, 2) parallel to 2x + 5y − z = 7.
7. Where does the line x = 1 + t, y = 2t, z = 4 − t meet the plane x + y + z = 11?
8. Find the distance from (1, −2, 3) to the plane 2x − y + 2z = 4.
9. Are the planes x − 2y + 3z = 1 and −2x + 4y − 6z = 5 parallel? Are they the same plane?
10. Find the angle between the planes x + y = 3 and y + z = 1.

4(x - 2) - 1(y - 1) + 5(z + 3) = 0

4x - 8 - y + 1 + 5z + 15 = 0

4x - y + 5z + 8 = 0

4x - y + 5z = -8

\overrightarrow{AB} = \langle 1, -1, 1 \rangle, \quad \overrightarrow{AC} = \langle -1, 2, 1 \rangle

(-1) \cdot 1 - 1 \cdot 2 = -3

-(1 \cdot 1 - 1 \cdot (-1)) = -2

1 \cdot 2 - (-1) \cdot (-1) = 1

-3(x - 1) - 2(y - 1) + 1(z - 0) = 0

-3x + 3 - 2y + 2 + z = 0

-3x - 2y + z = -5

3x + 2y - z = 5

4x = 20

x = 5

2y = 20

y = 10

5z = 20

z = 4

2 \cdot 1 - 3 + 4 \cdot (-2) = 2 - 3 - 8

= -9

2(x - 3) + 5(y + 1) - 1(z - 2) = 0

2x - 6 + 5y + 5 - z + 2 = 0

2x + 5y - z + 1 = 0

2x + 5y - z = -1

(1 + t) + 2t + (4 - t) = 11

5 + 2t = 11

2t = 6

t = 3

(1 + 3, \; 2 \cdot 3, \; 4 - 3) = (4, 6, 1)

2 \cdot 1 - (-2) + 2 \cdot 3 - 4 = 2 + 2 + 6 - 4

= 6

\sqrt{4 + 1 + 4} = 3

D = \dfrac{|6|}{3} = 2

\langle -2, 4, -6 \rangle = -2 \langle 1, -2, 3 \rangle

-2x + 4y - 6z = -2

-2 \ne 5

1 \cdot 0 + 1 \cdot 1 + 0 \cdot 1 = 1

\cos\theta = \dfrac{1}{\sqrt{2} \cdot \sqrt{2}} = \dfrac{1}{2}

\theta = 60^\circ

Equation of a plane FAQ

How do you find the equation of a plane from three points?

Subtract one point from the other two to get two vectors in the plane. Their cross product is a normal vector n. Then use n and any of the three points in a(x − x₀) + b(y − y₀) + c(z − z₀) = 0.

What is a normal vector to a plane?

It is a nonzero vector perpendicular to the plane. In ax + by + cz = d, the vector ⟨a, b, c⟩ is normal to the plane.

How do you find the distance from a point to a plane?

Substitute the point into ax + by + cz − d, take the absolute value, and divide by √(a² + b² + c²).

How can you tell if two planes are parallel?

Compare their normal vectors. If one is a scalar multiple of the other, the planes are parallel (or the same plane, if the whole equations are multiples).

Summary

• A plane is set by a point (x₀, y₀, z₀) and a normal vector n = ⟨a, b, c⟩.
• Point-normal form: a(x − x₀) + b(y − y₀) + c(z − z₀) = 0, or ax + by + cz = d.
• For three points, n = AB × AC.
• Distance from a point to a plane: |ax₁ + by₁ + cz₁ − d| ÷ √(a² + b² + c²).
• Parallel planes have parallel normals; the angle between planes is the angle between normals.

Previous lesson: Equations of Lines in 3D

Next lesson: Quadric Surfaces

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