How to Find Intersection of Two Planes
When working in three‑dimensional geometry, the intersection of two planes is a fundamental concept that appears in engineering, computer graphics, physics, and many applied mathematics problems. Now, understanding how to determine whether two planes intersect, and if so, what their line of intersection looks like, provides a solid foundation for solving systems of linear equations and visualizing spatial relationships. This guide walks you through the theory, the step‑by‑step procedure, and practical examples so you can confidently find the intersection of any pair of planes And that's really what it comes down to. Practical, not theoretical..
1. What Is a Plane in 3‑D Space?
A plane can be described by a linear equation of the form
[ Ax + By + Cz = D ]
where ((A, B, C)) is a normal vector (\mathbf{n}) perpendicular to the plane, and (D) is a constant that shifts the plane away from the origin. Any point ((x, y, z)) that satisfies the equation lies on the plane That's the part that actually makes a difference. And it works..
Two planes are given by
[ \begin{aligned} \mathbf{n}_1 \cdot \mathbf{r} &= D_1 \quad\text{with}\quad \mathbf{n}_1 = \langle A_1, B_1, C_1\rangle \ \mathbf{n}_2 \cdot \mathbf{r} &= D_2 \quad\text{with}\quad \mathbf{n}_2 = \langle A_2, B_2, C_2\rangle \end{aligned} ]
where (\mathbf{r} = \langle x, y, z\rangle) is the position vector of a generic point Less friction, more output..
2. Possible Outcomes When Two Planes Meet
Before diving into calculations, it helps to know the three geometric possibilities:
| Relationship | Condition | Result |
|---|---|---|
| Parallel and distinct | (\mathbf{n}_1) is a scalar multiple of (\mathbf{n}_2) and (D_1 \neq D_2) (scaled accordingly) | No intersection (empty set) |
| Coincident | (\mathbf{n}_1) is a scalar multiple of (\mathbf{n}_2) and the constants satisfy the same scaling ((D_1 = k D_2)) | Infinite intersection – the planes are the same |
| Intersecting in a line | (\mathbf{n}_1) and (\mathbf{n}_2) are not parallel | A unique line of intersection |
The focus of this article is the third case: finding that line.
3. Step‑by‑Step Procedure to Find the Intersection Line
The intersection line can be expressed in parametric form
[ \mathbf{r}(t) = \mathbf{r}_0 + t,\mathbf{d} ]
where
- (\mathbf{d}) – direction vector of the line (parallel to both planes)
- (\mathbf{r}_0) – a specific point that lies on both planes
Step 1: Compute the Direction Vector (\mathbf{d})
The direction vector is orthogonal to both normal vectors, so it is given by the cross product:
[ \mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 ]
If (\mathbf{d} = \mathbf{0}), the normals are parallel → either no intersection or coincident planes (handle separately).
Step 2: Find a Point (\mathbf{r}_0) on Both Planes
To obtain a concrete point, set one coordinate to a convenient value (often zero) and solve the resulting 2×2 linear system for the other two coordinates.
Common strategies:
- Set (z = 0) → solve for (x) and (y) from
[ \begin{cases} A_1 x + B_1 y = D_1 \ A_2 x + B_2 y = D_2 \end{cases} ] - If the above system is singular (determinant zero), try setting (x = 0) or (y = 0) instead.
Solve using substitution, elimination, or Cramer’s rule. The solution ((x_0, y_0, z_0)) gives (\mathbf{r}_0 = \langle x_0, y_0, z_0\rangle) Small thing, real impact..
Step 3: Write the Parametric Equation
Combine the point and direction:
[ \boxed{; \mathbf{r}(t) = \langle x_0, y_0, z_0\rangle + t,\langle d_x, d_y, d_z\rangle ;} ]
where (\langle d_x, d_y, d_z\rangle = \mathbf{d}) That's the whole idea..
Step 4: (Optional) Convert to Symmetric Form
If none of the components of (\mathbf{d}) are zero, you can write:
[ \frac{x - x_0}{d_x} = \frac{y - y_0}{d_y} = \frac{z - z_0}{d_z} ]
If a component is zero, the corresponding fraction is omitted and the equality is expressed as a constant (e.g., (x = x_0) when (d_x = 0)).
4. Worked Example
Problem: Find the intersection of the planes
[ \begin{aligned} \Pi_1:&; 2x - y + 3z = 7 \ \Pi_2:&; -x + 4y - z = -2 \end{aligned} ]
Step 1: Direction Vector
[ \mathbf{n}_1 = \langle 2, -1, 3\rangle,\qquad \mathbf{n}_2 = \langle -1, 4, -1\rangle ]
[ \mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\ 2 & -1 & 3\ -1 & 4 & -1 \end{vmatrix} = \mathbf{i}((-1)(-1)-3\cdot4) - \mathbf{j}(2\cdot(-1)-3\cdot(-1)) + \mathbf{k}(2\cdot4-(-1)\cdot(-1)) ]
[ = \mathbf{i}(1-12) - \mathbf{j}(-2+3) + \mathbf{k}(8-1) = \langle -11, -1, 7\rangle ]
So (\mathbf{d} = \langle -11, -1, 7\rangle) (non‑zero → planes are not parallel).
Step 2: Find a Point
Set (z = 0) and solve:
[ \begin{cases} 2x - y = 7 \ -x + 4y = -2 \end{cases} ]
From the first equation, (y = 2x - 7). Substitute into the second:
[ -x + 4(2x - 7) = -2 \ -x + 8x - 28 = -2 \ 7x = 26 \ x = \frac{26}{7} ]
Then (y = 2\left(\frac{26}{7}\right) - 7 = \frac{52}{7} - \frac{49}{7} = \frac{3}{7}).
Thus (\mathbf{r}_0 = \left\langle \frac{26}{7}, \frac{3}{7}, 0\right\rangle) Not complicated — just consistent..
Step 3: Parametric Form
[ \boxed{; \mathbf{r}(
