Line of Intersection of Two Planes Calculator
Find where two planes meet. Enter both equations and get the parametric line, its direction vector from the cross product, and the angle between them.
Line of Intersection of Two Planes Calculator
Background.
Two planes in three-dimensional space that are not parallel meet in a straight line — never in a single point, which is the first thing to get straight when moving up from two dimensions. This calculator takes both plane equations in the standard form a·x + b·y + c·z = d and returns that line as a parametric equation, along with its direction vector, a specific point it passes through, and the acute angle between the two planes.
The direction of the line comes from a cross product. A plane's coefficients (a, b, c) are its normal vector, the direction perpendicular to the surface. The line of intersection lies inside both planes at once, so it must be perpendicular to both normals — and the cross product n₁ × n₂ is by definition perpendicular to both. That single observation is the whole method for the direction, and it also explains the degenerate case: when the normals are parallel the cross product is the zero vector, which is exactly when the planes are parallel and there is no line to find.
Finding a point on the line takes one more step. The direction alone fixes the line's orientation but not its position, so the calculator pins one coordinate to zero and solves the remaining two-by-two system for the other two. Which coordinate gets zeroed is chosen deliberately: the one whose direction component is largest in magnitude, because that leaves the best-conditioned system to solve. A consequence worth stating plainly is that the point returned is not canonical. Any point on the line is equally valid, so your textbook's answer may show a different point and still be correct — check by substituting into both plane equations rather than by comparing points.
The direction vector is likewise reported exactly as the cross product computes it, neither scaled to unit length nor reduced to lowest terms. Any non-zero multiple of it names the same line, so a direction of (4, 2, −6) and a direction of (2, 1, −3) describe identical lines. Leaving the raw values visible lets you check the cross product by hand.
When the two planes are parallel or identical the calculator raises a specific error naming which case you have, rather than returning zeros. Parallel planes with different constants share no points at all; identical planes share every point of the plane rather than a line, and the calculator detects that even when the two equations are written at different scales, such as x + y + z = 6 and 2x + 2y + 2z = 12.
What is line of intersection of two planes calculator?
The line of intersection of two planes is the set of points lying on both. In three dimensions two distinct non-parallel planes always share exactly such a line: the system of two linear equations in three unknowns is underdetermined by one degree of freedom, and that free parameter is the t of the parametric equation. The line's direction is n₁ × n₂, the cross product of the planes' normal vectors, which is perpendicular to both normals and therefore parallel to both planes. The angle between two planes, called the dihedral angle, is defined as the angle between their normals, conventionally reported as the acute one.
How to use this calculator.
- Write both planes in the form a·x + b·y + c·z = d. If yours is written as a normal vector and a point, expand it first: a, b and c are the normal's components, and d is the normal dotted with the point.
- Enter the four numbers for each plane. At least one of a, b and c must be non-zero for each — all three zero is not a plane.
- Read the parametric equations at the top. Substituting any real number for t gives a point on the line of intersection.
- Check the answer by substituting the reported point into both original equations; it should satisfy both exactly. Then add the direction vector to it and check again — that second point must also satisfy both.
- If your textbook shows a different point on the line, both can be right. Any point on the line is a valid base point, so compare by substitution rather than by matching coordinates.
The formula.
The line of intersection lies in both planes, so it is perpendicular to both normal vectors. The cross product n₁ × n₂ is perpendicular to both by construction, so it is the line's direction. Written out, its components are b₁c₂ − c₁b₂ for x, c₁a₂ − a₁c₂ for y, and a₁b₂ − b₁a₂ for z. The middle term is the one most often written with the wrong sign by hand, because the cofactor expansion alternates signs; writing it as c₁a₂ − a₁c₂ rather than as the negative of a₁c₂ − c₁a₂ removes the trap.
A direction is not yet a line. To fix the position, the calculator sets one coordinate to zero and solves the two remaining equations for the other two unknowns. Setting z = 0, for instance, reduces the problem to a₁x + b₁y = d₁ and a₂x + b₂y = d₂, an ordinary two-by-two system whose determinant is a₁b₂ − a₂b₁ — which is precisely the z-component of the cross product already computed. That is why the coordinate to zero out is chosen as the one whose direction component is largest in magnitude: it guarantees a non-zero determinant and picks the best-conditioned of the three possible systems.
The resulting point is one point on the line, not the point. Choosing a different coordinate to zero, or simply walking along the line, gives a different base point describing the same line. Two answers that look different are both correct if each point satisfies both plane equations and each direction is a non-zero multiple of the other.
The magnitude of the direction vector is worth reading. Because ‖n₁ × n₂‖ = ‖n₁‖‖n₂‖ sin θ, it shrinks toward zero as the planes approach parallel, and it is exactly zero when they are. That is the degeneracy test the calculator applies: a zero cross product means no unique line exists. It then distinguishes the two sub-cases. If the constants scale by the same factor as the coefficients, the two equations describe the same plane and share all of its points; otherwise the planes are parallel and distinct and share none.
The angle between the planes is the angle between their normals, cos θ = |n₁·n₂| ⁄ (‖n₁‖‖n₂‖). The absolute value makes the result acute, in the range 0° to 90°. It also makes the answer independent of how each equation happens to be signed: multiplying a plane equation by −1 reverses its normal but obviously does not move the plane, and without the absolute value that harmless rewriting would return the supplementary angle instead.
All arithmetic runs in arbitrary-precision decimal, and rounding to ten decimal places happens once, at the point the answer is returned. Negative zero is normalised, so a coordinate of zero always displays as 0.
A worked example.
Find where x + y + z = 6 meets 2x − y + z = 3. The normals are n₁ = (1, 1, 1) and n₂ = (2, −1, 1). Their cross product gives the direction: the x-component is b₁c₂ − c₁b₂ = 1·1 − 1·(−1) = 2, the y-component is c₁a₂ − a₁c₂ = 1·2 − 1·1 = 1, and the z-component is a₁b₂ − b₁a₂ = 1·(−1) − 1·2 = −3. So the line runs along v = (2, 1, −3), with length √(4 + 1 + 9) = √14 ≈ 3.7416573868. The largest direction component in magnitude is the z one, so the calculator sets z = 0 and solves what is left: x + y = 6 and 2x − y = 3. Adding the two equations eliminates y and gives 3x = 9, so x = 3, and then y = 6 − 3 = 3. The point is (3, 3, 0). Check it in both originals: 3 + 3 + 0 = 6, which is d₁, and 2(3) − 3 + 0 = 3, which is d₂. Both hold exactly. Check the direction too, by confirming it is perpendicular to each normal: v·n₁ = 2 + 1 − 3 = 0 and v·n₂ = 4 − 1 − 3 = 0. Both zero, as they must be. The parametric form is therefore x = 3 + 2t, y = 3 + t, z = 0 − 3t. Setting t = 1 gives the point (5, 4, −3), and substituting that into the originals gives 5 + 4 − 3 = 6 and 10 − 4 − 3 = 3 — both still satisfied, which confirms the direction as well as the base point. The angle between the planes comes from the normals. Their dot product is 1(2) + 1(−1) + 1(1) = 2, and the magnitudes are √3 and √6, so cos θ = 2 ⁄ √18 ≈ 0.4714045208 and θ ≈ 61.8744942979°. The planes are neither parallel nor perpendicular. If you worked this by hand and got the point (0, 1.5, 4.5) instead, you are not wrong — that is what you get by setting x = 0 rather than z = 0, and it is a different point on the same line. It satisfies both equations (0 + 1.5 + 4.5 = 6 and 0 − 1.5 + 4.5 = 3), and it is the point you reach from (3, 3, 0) by taking t = −1.5.
Frequently asked questions.
Why do two planes intersect in a line rather than a point?
Why is the direction the cross product of the normals?
My textbook shows a different point on the line. Which is right?
Why is the direction vector not reduced or normalised?
What happens if the two planes are parallel?
References& sources.
- [1]OpenStax. Calculus Volume 3, §2.5 "Equations of Lines and Planes in Space" — the scalar and parametric equations of a line, the standard-form equation of a plane, the cross product of normals as the direction of the intersection line, and the dihedral angle. Open access.
- [2]Dawkins, Paul. "Equations of Planes." Paul's Online Math Notes, Lamar University Department of Mathematics — the normal-vector form of a plane and the parallel/orthogonal tests used here. Open access.
- [3]Weisstein, Eric W. "Plane-Plane Intersection." MathWorld — Wolfram Research. Independent statement that the intersection line's direction is the cross product of the two plane normals. Retrieved 2026-07-29.
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