Equations of Planes in 3D

Concept Explanation

Geometric Intuition & Vector Derivation

To define a line in 2D, you need a point and a direction slope. In 3D space, however, a single direction vector cannot define a unique plane because infinitely many planes can rotate around that direction. Instead, a plane is locked in space by specifying a direction that is completely perpendicular to its entire surface: the normal vector $\vec{n} = \langle a, b, c \rangle$.

Let $P_0(x_0, y_0, z_0)$ be a fixed base point on the plane with position vector $\vec{r}_0 = \langle x_0, y_0, z_0 \rangle$, and let $P(x, y, z)$ be any arbitrary point in space with position vector $\vec{r} = \langle x, y, z \rangle$. The displacement vector connecting $P_0$ to $P$ is:

The point $P(x, y, z)$ lies on the plane if and only if $\vec{P_0P}$ is completely parallel to the plane surface, which means it must be strictly perpendicular to the normal vector $\vec{n}$. Applying the Dot Product orthogonality condition:

Expanding the dot product gives the Point-Normal Form:

                   ^ n = <a, b, c> (Perpendicular Normal)
                   |
                   |
         +---------|-------------------+
        /          |                  /
       /           | P_0(x_0, y_0, z_0)
      /            +---------> P(x, y, z)
     /              r - r_0           /  Plane surface: n . (r - r_0) = 0
    +--------------------------------+

From Point-Normal to General Form

Distributing the coefficients $a, b, c$ across the parentheses yields:

Setting the scalar constant $d = ax_0 + by_0 + cz_0$ produces the standard General Form:

Determining a Plane from Three Non-Collinear Points

Three distinct, non-collinear points $A, B, C$ uniquely determine a plane in 3D:

1. Construct two independent in-plane displacement vectors sharing a common vertex:
   $$\vec{u} = \vec{AB} = \langle x_B - x_A, y_B - y_A, z_B - z_A \rangle$$
   $$\vec{v} = \vec{AC} = \langle x_C - x_A, y_C - y_A, z_C - z_A \rangle$$
2. Compute the Cross Product $\vec{u} \times \vec{v}$. By definition, the cross product produces a vector that is simultaneously orthogonal to both $\vec{u}$ and $\vec{v}$, serving as the plane's normal vector:
   $$\vec{n} = \vec{AB} \times \vec{AC}$$
3. Substitute $\vec{n}$ and any of the three points (e.g., $A$) into the point-normal form.

Parallel and Orthogonal Planes

The spatial orientation of a plane is entirely governed by its normal vector:

• Parallel Planes: Two planes with normal vectors $\vec{n}_1$ and $\vec{n}_2$ are parallel if and only if their normal vectors are scalar multiples:
  $$\vec{n}_1 = k\vec{n}_2, \quad k \neq 0$$
• Orthogonal (Perpendicular) Planes: Two planes are perpendicular if their normal vectors are orthogonal:
  $$\vec{n}_1 \cdot \vec{n}_2 = 0$$
• Angle Between Two Planes: The dihedral angle $\theta \in [0, \frac{\pi}{2}]$ between two intersecting planes is the acute angle between their normal vectors:
  $$\cos\theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{\|\vec{n}_1\| \|\vec{n}_2\|}$$

Geometric Meaning of Zero Coefficients

When one or two components of $\vec{n} = \langle a, b, c \rangle$ are zero, the plane adopts a special orientation relative to the Cartesian coordinate axes:

• If $a = 0$, the equation is $by + cz = d$. The normal vector $\vec{n} = \langle 0, b, c \rangle$ is perpendicular to the $x$-axis, meaning the plane itself is parallel to the $x$-axis.
• If $a = b = 0$, the equation simplifies to $cz = d \implies z = \frac{d}{c}$, which represents a horizontal plane parallel to the $xy$-coordinate plane.

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Engineering Applications

• Robotics & Safe Workspaces: Defining virtual planar boundary barriers in robotic arm control loops to prevent tooling end-effectors from colliding with conveyors or human workstations.
• Civil & Structural BIM Modeling: Establishing 3D Building Information Modeling (BIM) datum planes for structural floor slabs, sloping roof trusses, and precast curtain wall facades.
• Computer Vision & LiDAR Point Clouds: Segmenting flat surfaces (walls, roadways, ramps) from raw 3D LiDAR point cloud scans using RANSAC plane-fitting algorithms.
• Aerospace Control Surfaces: Defining planar hinge references and aerodynamic deflector orientations on supersonic control fins.

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Example Problems

Example 1: Point-Normal to General Form with a Missing Variable

Find the general equation of the plane that passes through $P(2, 5, 6)$ and has normal vector $\vec{n} = \langle 0, 2, 6 \rangle = 2\hat{j} + 6\hat{k}$.

Solution

1. Identify the Components:
   • Normal vector components: $a = 0, b = 2, c = 6$.
   • Known base point: $x_0 = 2, y_0 = 5, z_0 = 6$.
2. Substitute into Point-Normal Form:
   $$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$$
   $$0(x - 2) + 2(y - 5) + 6(z - 6) = 0$$
3. Distribute and Simplify:
   $$2y - 10 + 6z - 36 = 0 \implies 2y + 6z = 46$$
4. Divide by Common Factor 2:
   $$y + 3z = 23$$
   Geometric Note: Because $x$ is absent ($a = 0$), the plane is parallel to the $x$-axis.

Example 2: Plane Determined by Three Points

Find the equation of the plane passing through $A(1, 2, -1)$, $B(3, -1, 2)$, and $C(0, 1, 1)$.

Solution

1. Form Two In-Plane Displacement Vectors:
   $$\vec{u} = \vec{AB} = \langle 3 - 1, -1 - 2, 2 - (-1) \rangle = \langle 2, -3, 3 \rangle$$
   $$\vec{v} = \vec{AC} = \langle 0 - 1, 1 - 2, 1 - (-1) \rangle = \langle -1, -1, 2 \rangle$$
2. Compute the Normal Vector via Cross Product:
   $$\vec{n} = \vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 3 \\ -1 & -1 & 2 \end{vmatrix}$$
   $$\vec{n} = \hat{i}[(-3)(2) - (3)(-1)] - \hat{j}[(2)(2) - (3)(-1)] + \hat{k}[(2)(-1) - (-3)(-1)]$$
   $$\vec{n} = \hat{i}[-6 + 3] - \hat{j}[4 + 3] + \hat{k}[-2 - 3] = -3\hat{i} - 7\hat{j} - 5\hat{k} = \langle -3, -7, -5 \rangle$$
   Multiply by $-1$ to obtain an equivalent positive normal vector: $\vec{n} = \langle 3, 7, 5 \rangle$.
3. Apply Point-Normal Form Using Point $C(0, 1, 1)$:
   $$3(x - 0) + 7(y - 1) + 5(z - 1) = 0$$
   $$3x + 7y - 7 + 5z - 5 = 0$$
   $$3x + 7y + 5z = 12$$
4. Verification with Points $A$ and $B$:
   • Point $A(1, 2, -1)$: $3(1) + 7(2) + 5(-1) = 3 + 14 - 5 = 12$. (Verified)
   • Point $B(3, -1, 2)$: $3(3) + 7(-1) + 5(2) = 9 - 7 + 10 = 12$. (Verified)

Example 3: Constructing a Parallel Plane

Find the equation of the plane passing through $P(-2, 3, 1)$ that is parallel to the plane $4x - 2y + 5z = 19$.

Solution

1. Extract the Normal Vector: From $4x - 2y + 5z = 19$, the normal vector is $\vec{n} = \langle 4, -2, 5 \rangle$.
2. Apply Parallelism: Parallel planes share the same normal vector orientation. Thus, use $\vec{n} = \langle 4, -2, 5 \rangle$.
3. Construct the Equation with Point $P(-2, 3, 1)$:
   $$4(x - (-2)) - 2(y - 3) + 5(z - 1) = 0$$
   $$4(x + 2) - 2(y - 3) + 5(z - 1) = 0$$
   $$4x + 8 - 2y + 6 + 5z - 5 = 0$$
   $$4x - 2y + 5z + 9 = 0 \implies 4x - 2y + 5z = -9$$

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Connections & References

• Parent Concepts: Vectors in 2D and 3D | Dot and Cross Products
• Related Notes: Equations of Lines in 3D | Tangent Planes | Cross Product | Dot Product
• Course Sessions: Calc-3 Session 10