Lines and Planes in Space: Learn It 4

Equations for a Plane

Just as a line is uniquely determined by two points, a plane is uniquely determined by three points that don’t all lie on the same line. There are several ways to characterize a plane mathematically.

A plane can be determined by:

  • Three non-collinear points (the most basic definition)
  • Two distinct, intersecting lines (exactly one plane contains both)
  • One line and a point not on the line (the line and point determine the plane)
  • A point and a normal vector

The most powerful way to describe a plane uses a normal vector—a vector perpendicular to the plane. Here’s the key insight:

Imagine a vector [latex]\mathbf{n}[/latex] pointing perpendicular to a plane, and a fixed point [latex]P[/latex] on the plane. Any other point [latex]Q[/latex] lies on the same plane if and only if the vector [latex]\overrightarrow{PQ}[/latex] is perpendicular to [latex]\mathbf{n}[/latex].

This figure is a parallelogram representing a plane. In the plane is a vector from point P to point Q. Perpendicular to the vector P Q is the vector n.
Figure 1. Given a point [latex]P[/latex] and vector [latex]\textbf n[/latex], the set of all points [latex]Q[/latex] with [latex]\overrightarrow{PQ}[/latex] orthogonal to [latex]\textbf n[/latex] forms a plane.

Since orthogonal vectors have dot product zero, this gives us our plane equation.

plane equations

Given point [latex]P = (x_0, y_0, z_0)[/latex] on the plane and normal vector [latex]\mathbf{n} = \langle a, b, c \rangle[/latex]:

  • Vector equation: [latex]\mathbf{n} \cdot \overrightarrow{PQ} = 0[/latex]
  • Scalar equation: [latex]a(x - x_0) + b(y - y_0) + c(z - z_0) = 0[/latex]
  • General form: [latex]ax + by + cz + d = 0[/latex] where [latex]d = -ax_0 - by_0 - cz_0[/latex]

To see how these forms connect, let’s expand the vector equation. If [latex]Q = (x,y,z)[/latex] is any point on the plane, then [latex]\overrightarrow{PQ} = \langle x-x_0, y-y_0, z-z_0 \rangle[/latex]. The condition that this vector is orthogonal to [latex]\mathbf{n}[/latex] gives us:

[latex]\begin{aligned} \mathbf{n} \cdot \overrightarrow{PQ} &= 0 \ \langle a,b,c \rangle \cdot \langle x-x_0, y-y_0, z-z_0 \rangle &= 0 \ a(x-x_0) + b(y-y_0) + c(z-z_0) &= 0 \end{aligned}[/latex]

Expanding this gives us [latex]ax + by + cz - ax_0 - by_0 - cz_0 = 0[/latex], which we can write as [latex]ax + by + cz + d = 0[/latex] where [latex]d = -ax_0 - by_0 - cz_0[/latex].

When given three points, you need to find a normal vector to write the plane equation. The cross product provides the solution.

Write an equation for the plane containing points [latex]P=(1,1,-2)[/latex], [latex]Q=(0,2,1)[/latex], and [latex]R=(-1,-1,0)[/latex] in both standard and general forms.

The same plane can be described by different equations depending on which point and normal vector you choose. Scalar multiples of a normal vector still point in the same direction, so [latex]\langle 8,-4,4 \rangle[/latex], [latex]\langle 2,-1,1 \rangle[/latex], and [latex]\langle -4,2,-2 \rangle[/latex] all describe the same plane orientation. [/Tip Box]

Connection to Cross Products: Finding a plane through three points always involves:

  1. Form two vectors from the three points
  2. Take their cross product to get a normal vector
  3. Use the normal vector and any of the three points to write the plane equation
Find an equation of the plane that passes through point [latex](1, 4, 3)[/latex] and contains the line given by [latex]x=\frac{y-1}2=z+1[/latex].

Distance from a Point to a Plane

Once you have a plane equation, you can find the shortest distance from any point to the plane. This distance is always measured along a line perpendicular to the plane.

This figure is the sketch of a parallelogram representing a plane. In the plane are points Q and R. there is a broken line from Q to R on the plane. There is a vector n out of the plane at point Q. Also, there is a vector labeled “R P” from point R to point P which is above the plane. This vector is perpendicular to the plane.
Figure 3. We want to find the shortest distance from point [latex]P[/latex] to the plane. Let point [latex]R[/latex] be the point in the plane such that, for any other point in the plane [latex]Q[/latex], [latex]||\overrightarrow{RP}||<||\overrightarrow{QP}||[/latex].

Just as you find the 2D distance from a point to a line by measuring perpendicular to the line, you find the 3D distance from a point to a plane by measuring perpendicular to the plane.

If [latex]R[/latex] is the closest point on the plane to [latex]P[/latex], then [latex]\overrightarrow{RP}[/latex] must be parallel to the normal vector [latex]\mathbf{n}[/latex]. For any other point [latex]Q[/latex] on the plane, the vector [latex]\overrightarrow{QP}[/latex] can be projected onto [latex]\mathbf{n}[/latex] to find [latex]\overrightarrow{RP}[/latex].

the distance between a plane and a point

For a plane with normal vector [latex]\mathbf{n}[/latex] passing through point [latex]Q[/latex], the distance [latex]d[/latex] from the plane to point [latex]P[/latex] (not on the plane) is:

[latex]d=\left|\left|\text{proj}_{\bf{n}}\overrightarrow{QP}\right|\right|=\left|\text{comp}_{\bf{n}}\overrightarrow{QP}\right|=\frac{|\overrightarrow{QP}\cdot{\bf{n}}|}{||{\bf{n}}||}[/latex].

Why the Absolute Value? Distance is always positive, but [latex]\overrightarrow{QP} \cdot \mathbf{n}[/latex] can be positive or negative depending on which side of the plane point [latex]P[/latex] lies. The absolute value ensures we get a positive distance
This formula is exactly the scalar projection of [latex]\overrightarrow{QP}[/latex] onto the normal vector [latex]\mathbf{n}[/latex]. The projection gives the component of [latex]\overrightarrow{QP}[/latex] in the direction perpendicular to the plane—which is precisely the shortest distance.

Find the distance between point [latex]P=(3, 1, 2)[/latex] and the plane given by [latex]x-2y+z=5[/latex] (see the following figure).

This figure is the 3-dimensional coordinate system. There is a point drawn at (3, 1, 2). The point is labeled “P(3, 1, 2).” There is a plane drawn. There is a perpendicular line from the plane to point P(3, 1, 2).
Figure 4. The distance between the plane given by [latex]x-2y+z=5[/latex] and the point [latex]P=(3, 1, 2)[/latex]