Free Line of Intersection of Two Planes Calculator

Plane 1

a₁x + b₁y + c₁z = d₁

Plane 2

a₂x + b₂y + c₂z = d₂

Intersection Status

Enter both plane equations
to find the line of intersection

Planes and Their Line of Intersection

In three-dimensional space, a plane is a flat, two-dimensional surface with zero curvature. When two such planes are not parallel, their intersection automatically forms a straight line of intersection — a fundamental result in spatial geometry. The Intersection of Two Planes calculator offers a fast way to derive this line, presenting it as a parametric line equation that can be used for further analysis in engineering, physics, or computer graphics.

Representing a Plane

A plane is fully described by a point and a normal vector that is perpendicular to the surface. If the point is P0=(x0,y0,z0)P_0 = (x_0, y_0, z_0) and the normal is n=⟨a,b,c⟩\mathbf{n} = \langle a, b, c \rangle, the Cartesian equation is:

a(x−x0)+b(y−y0)+c(z−z0)=0orax+by+cz=d,a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \quad \text{or} \quad a x + b y + c z = d,

where d=ax0+by0+cz0d = a x_0 + b y_0 + c z_0. The normal vector governs the plane's orientation, while the point determines its distance from the origin.

Intersection Geometry

Two planes can be:

  • Parallel (same normal direction) — no intersection line exists; they are either coincident or separated.
  • Intersecting (normal vectors are not parallel) — the intersection is a straight line.

The line belongs to both planes simultaneously and is essentially the set of all points that satisfy both plane equations.

Computing the Intersection Line – Parametric Method

The line of intersection can be expressed in parametric form using three steps:

  1. Direction vector – Because the line lies in both planes, it is perpendicular to each plane's normal. The cross product of the normal vectors gives the line's direction:
r=n1×n2=∣ijka1b1c1a2b2c2∣.\mathbf{r} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}.
  1. Common point – Solve the two plane equations together (usually by setting one coordinate to zero) to find a point P0=(x0,y0,z0)P_0 = (x_0, y_0, z_0) that lies on both planes.

  2. Parametric equation – Combine the point and direction:

l(λ)=⟨x0,y0,z0⟩+λ⟨rx,ry,rz⟩,\mathbf{l}(\lambda) = \langle x_0, y_0, z_0 \rangle + \lambda \langle r_x, r_y, r_z \rangle,

which is equivalent to:

\begin{cases} x(\lambda) = x_0 + \lambda r_x,\$$2pt] y(\lambda) = y_0 + \lambda r_y,\$$2pt] z(\lambda) = z_0 + \lambda r_z. \end{cases}

The parameter λ\lambda can take any real value; each value yields a corresponding point on the intersection line.

Example

Take these two planes:

A:−2x+3y+4z=−1,B:2x−y−3z=2.\begin{aligned} A &: -2x + 3y + 4z = -1,\\ B &: 2x - y - 3z = 2. \end{aligned}

Normals: n1=⟨−2,3,4⟩\mathbf{n}_1 = \langle -2, 3, 4 \rangle, n2=⟨2,−1,−3⟩\mathbf{n}_2 = \langle 2, -1, -3 \rangle.

Cross product:

r=⟨(−3)(−3)−(4)(−1),  (4)(2)−(−2)(−3),  (−2)(−1)−(3)(2)⟩=⟨−5,2,−4⟩.\mathbf{r} = \langle (-3)(-3) - (4)(-1),\; (4)(2) - (-2)(-3),\; (-2)(-1) - (3)(2) \rangle = \langle -5, 2, -4 \rangle.

To get a common point, set x=0x = 0 and solve the system:

{3y+4z=−1,−y−3z=2.\begin{cases} 3y + 4z = -1,\\ -y - 3z = 2. \end{cases}

This yields y=1y = 1, z=−1z = -1, so P0=(0,1,−1)P_0 = (0, 1, -1). The parametric line is:

l(λ)=⟨0,1,−1⟩+λ⟨−5,2,−4⟩.\mathbf{l}(\lambda) = \langle 0, 1, -1 \rangle + \lambda \langle -5, 2, -4 \rangle.

Intersection Line in Symmetric Form

The parametric equation can be converted into a symmetric form by eliminating λ\lambda, provided that no component of r\mathbf{r} is zero:

x−x0rx=y−y0ry=z−z0rz.\frac{x - x_0}{r_x} = \frac{y - y_0}{r_y} = \frac{z - z_0}{r_z}.

Alternatively, the intersection can be found directly by elimination without using a parameter.

Example Using Elimination

Consider:

2x−4z=−1,x−2y+z=2.\begin{aligned} &2x - 4z = -1,\\ &x - 2y + z = 2. \end{aligned}

From the first equation, isolate zz:

z=12x+14.z = \frac{1}{2}x + \frac{1}{4}.

Substitute into the second:

x−2y+12x+14=2  ⟹  32x−2y=74.x - 2y + \frac{1}{2}x + \frac{1}{4} = 2 \;\Longrightarrow\; \frac{3}{2}x - 2y = \frac{7}{4}.

Solve for xx:

x=43y+76.x = \frac{4}{3}y + \frac{7}{6}.

Also from the first equation, express xx in terms of zz:

x=2z+12.x = 2z + \frac{1}{2}.

Thus the line is described symmetrically by the two equations:

x=43y+76=2z+12.x = \frac{4}{3}y + \frac{7}{6} = 2z + \frac{1}{2}.

Using the Calculator

The Line of Intersection of Two Planes Calculator simplifies every step. You enter the coefficients (a,b,c,d)(a,b,c,d) for each plane in the standard form ax+by+cz=dax+by+cz = d. The tool automatically computes the cross product of the normals to obtain the direction vector, finds a convenient common point, and displays the result as a parametric equation. It also alerts you when the planes are parallel—meaning no intersection exists.

FAQ

1. How does the calculator determine the line of intersection of two planes?

It computes the cross product of the two planes' normal vectors to get the direction vector, then solves the plane equations (often by setting one coordinate to zero) to find a common point, and finally combines these into a parametric equation.

2. What happens if the two planes are parallel?

If the normal vectors are parallel (scalar multiples), the planes are either coincident or separate; no unique intersection line exists. The calculator detects this condition and notifies the user.

3. Can I obtain the symmetric (Cartesian) form of the intersection line from the calculator output?

Yes. The parametric output can be converted to symmetric form by eliminating the parameter λ, resulting in (x − x₀)/r_x = (y − y₀)/r_y = (z − z₀)/r_z, provided none of the direction components is zero.

4. Why is the cross product of the normal vectors used?

The cross product yields a vector perpendicular to both normals. Since the intersection line lies in both planes, it must be perpendicular to each plane's normal, so this cross product gives the line's exact direction.

5. Do I always need to set one variable to zero to find the common point?

Setting one variable to zero is a common technique to simplify the system of two equations, but any method that yields a point satisfying both plane equations works. The calculator handles this step automatically.

How to Use

  1. Enter the coefficients of the first plane equation in the form a₁x + b₁y + c₁z = d₁.
  2. Enter the coefficients of the second plane equation in the form a₂x + b₂y + c₂z = d₂.
  3. The parametric equation of the line of intersection is calculated instantly, showing the direction vector and common point.