Source: MATH 163 Class Notes, Section 12.5
Difficulty: Intermediate | Prerequisites: Plane equations (standard and vector form), cross products, dot products, parametric line equations. Ideally read the Distance and Plane Equations study notes first.
Once you know how to write plane equations, the natural next questions are: how do planes relate to each other and to lines? This section covers the angle between two planes, the line where two planes intersect, and the point where a line pierces a plane. These are the geometric building blocks for everything from surface normals in physics to intersection testing in computer graphics. You should already be comfortable writing a plane equation in standard form and a line in parametric form before working through this material.
The angle between two planes comes from the dot product of their normals. The line of intersection of two planes has a direction given by the cross product of the normals, and a starting point found by solving the two plane equations simultaneously. To find where a line hits a plane, substitute the parametric equations into the plane equation and solve for t.
Angle between two planes
The angle θ between two planes is the angle between their normal vectors. Computed using the dot product formula. Think of it as measuring how "tilted" one plane is relative to the other.
Line of intersection
The line where two non-parallel planes meet. Every point on this line satisfies both plane equations simultaneously. In simple terms, it is the crease where the two planes fold together.
Parametric form
x(t) = x_0 + at, y(t) = y_0 + bt, z(t) = z_0 + ct. Each coordinate written as a function of one parameter t. This is the form you use when substituting a line into a plane equation.
Symmetric form
(x - x_0)/a = (y - y_0)/b = (z - z_0)/c. Derived from parametric form by solving each equation for t and setting them equal. Think of it as a compact notation that eliminates the parameter.
Vector form of a line
r(t) = P_0 + vt, where P_0 is a point on the line and v is its direction vector. The most concise way to write a line in 3D.
The angle between two planes equals the angle between their normal vectors. Use the standard dot product angle formula.
Read off the normal vectors from the standard-form equations. For ax + by + cz = d, the normal is ⟨a, b, c⟩.
Apply θ = cos⁻¹( (n_1 · n_2) / (|n_1| |n_2|) ).
Worked example: Planes x + y + z = 1 and x - 2y + 3z = 1.
n_1 = ⟨1, 1, 1⟩, n_2 = ⟨1, -2, 3⟩.
n_1 · n_2 = 1 - 2 + 3 = 2.
|n_1| = √3, |n_2| = √14.
θ = cos⁻¹(2 / (√3 · √14)) = cos⁻¹(2 / √42) ≈ 72° ≈ 1.26 rad.
Note: this formula can return an obtuse angle. If the problem asks for the acute angle between the planes, take 180° minus the result when θ > 90°.
When two planes are not parallel, they meet along a line. Finding that line requires two things: a direction vector and a starting point.
Finding the direction vector:
The line lies in both planes, so it is perpendicular to both normal vectors.
Take the cross product of the two normals: direction = n_1 × n_2.
Example: n_1 = ⟨1, 1, 1⟩, n_2 = ⟨1, -2, 3⟩. Direction = ⟨1, 1, 1⟩ × ⟨1, -2, 3⟩ = ⟨5, -2, -3⟩.
Finding a starting point:
The starting point must satisfy both plane equations at the same time.
Set one variable to an arbitrary value (typically z = 0) to reduce to a system of two equations in two unknowns.
Example: with z = 0, the system becomes x + y = 1 and x - 2y = 1. Solving gives x = 1, y = 0. So the starting point is (1, 0, 0).
Writing the line in all three forms:
Vector form: r(t) = ⟨1, 0, 0⟩ + ⟨5, -2, -3⟩t.
Parametric form: x(t) = 1 + 5t, y(t) = -2t, z(t) = -3t.
Symmetric form: (x - 1)/5 = y/(-2) = z/(-3).
The choice of z = 0 is arbitrary. Any value works, and different choices give different but equivalent starting points on the same line.
To find where a line hits a plane, substitute the parametric equations of the line into the plane equation and solve for t. Then plug t back in to get the coordinates.
Worked example: Find the intersection of the plane 5x - 4y + 2z = 3 and the line x(t) = 3t + 1, y(t) = -2t, z(t) = 5t - 4.
Substitute into the plane equation: 5(3t + 1) - 4(-2t) + 2(5t - 4) = 3.
Expand: 15t + 5 + 8t + 10t - 8 = 3.
Simplify: 33t - 3 = 3, so 33t = 6, giving t = 2/11.
Plug back in: x(2/11) = 17/11, y(2/11) = -4/11, z(2/11) = -34/11.
Intersection point: (17/11, -4/11, -34/11).
If solving for t gives a contradiction (no solution), the line is parallel to the plane and does not intersect it. If every value of t satisfies the equation, the line lies entirely within the plane.
Formula | What it gives you |
|---|---|
θ = cos⁻¹( (n_1 · n_2) / (|n_1| |n_2|) ) | Angle between two planes |
Direction = n_1 × n_2 | Direction vector for the line of intersection |
Substitute x(t), y(t), z(t) into Ax + By + Cz = D, solve for t | The parameter value where a line meets a plane |
Students sometimes use the dot product of the normals to find the direction of the intersection line. The dot product gives a scalar, not a vector. You need the cross product for the direction.
When finding a starting point on the intersection line, students try to solve three unknowns with only two equations without setting one variable to a constant. You must fix one variable (usually z = 0) to make the system solvable.
Students forget that the angle formula can return an obtuse angle. Two planes can meet at either an acute or obtuse angle depending on the orientation of the normals, so check whether the problem asks for the acute angle.
When substituting a line into a plane, students occasionally substitute the direction vector components instead of the full parametric expressions. Always use x(t) = x_0 + at, not just a.
⚠️ "Find the line of intersection and express in vector, parametric, and symmetric form" is a classic multi-part exam question. Practise the full pipeline from normals to all three forms.
⚠️ Finding the angle between two planes is quick to compute but easy to lose marks on if you forget to check whether the problem wants the acute angle.
⚠️ Plane-line intersection is often combined with a follow-up question (distance from that point to another object, or checking whether a point lies on a given side of the plane).
⚠️ Know the special cases: if n_1 × n_2 = 0, the planes are parallel (no intersection line). If the substitution for a line-plane intersection gives 0 = 0, the line lies in the plane.
True or false: The direction of the line of intersection of two planes is found using the dot product of their normals. (False, use the cross product.)
Fill in the blank: To find a starting point on the intersection line, fix one variable (e.g. z = 0) and solve the remaining system of ____ equations in ____ unknowns. (two, two)
True or false: If the cross product of two planes' normal vectors is the zero vector, the planes intersect along a line. (False, the planes are parallel.)
Fill in the blank: To find where a line meets a plane, substitute the ____ equations into the plane equation. (parametric)
True or false: The angle between two planes is always acute. (False, the dot product formula can return an obtuse angle.)
Q: Find the angle between the planes 3x + y - z = 5 and x - 2y + 4z = 7.
A: n_1 = ⟨3, 1, -1⟩, n_2 = ⟨1, -2, 4⟩. Dot product: 3 - 2 - 4 = -3. |n_1| = √11, |n_2| = √21. θ = cos⁻¹(-3 / √231) ≈ 101.4°. The acute angle between the planes is 180° - 101.4° = 78.6°.
Q: Find the line of intersection of x + 2y - z = 3 and 2x - y + z = 1 in all three forms.
A: Direction = n_1 × n_2 = ⟨1, 2, -1⟩ × ⟨2, -1, 1⟩ = ⟨(2)(1) - (-1)(-1), (-1)(2) - (1)(1), (1)(-1) - (2)(2)⟩ = ⟨1, -3, -5⟩. Set z = 0: x + 2y = 3 and 2x - y = 1. From the second equation, y = 2x - 1. Substituting: x + 2(2x - 1) = 3, so 5x = 5, x = 1, y = 1. Starting point: (1, 1, 0). Vector form: r(t) = ⟨1, 1, 0⟩ + ⟨1, -3, -5⟩t. Parametric form: x = 1 + t, y = 1 - 3t, z = -5t. Symmetric form: (x - 1)/1 = (y - 1)/(-3) = z/(-5).
Q: Does the line x(t) = 2t, y(t) = t + 1, z(t) = -t + 3 intersect the plane x + y + z = 6? If so, where?
A: Substitute: 2t + (t + 1) + (-t + 3) = 6. This gives 2t + 4 = 6, so t = 1. Intersection point: (2, 2, 2).
Q: Show that the line x(t) = 1 + t, y(t) = 2 - t, z(t) = 3 + t is parallel to the plane x + y = 5.
A: The plane's normal is ⟨1, 1, 0⟩. The line's direction is ⟨1, -1, 1⟩. Dot product: 1 - 1 + 0 = 0. Since the direction is perpendicular to the normal, the line is parallel to the plane. Checking a point: 1 + 2 = 3 ≠ 5, so the line does not lie in the plane.
The angle between planes connects to the angle between vectors from earlier in the course (Section 12.3). The same cos⁻¹ formula applies, just with normal vectors instead of arbitrary ones.
Line-plane intersection is the foundation for parametric surface intersection problems in Chapter 16, where you test whether a parametric curve passes through a given surface.
The technique of setting one variable to zero and solving a 2×2 system appears again when finding critical points in optimisation (Chapter 14), so it is worth practising until it is second nature.
Angle between two planes formula, line of intersection of two planes, plane-line intersection point, parametric line intersection, symmetric form of a line, vector form of a line, cross product of normal vectors, dot product angle, parallel planes, perpendicular planes, MATH 163, Calculus III, Section 12.5, multivariable calculus geometry, 3D line equations