Line of Intersection of Two Planes Calculator

Created by Davide Borchia
Reviewed by Rijk de Wet
Last updated: Jun 05, 2023
Table of contents:

Calculating the line of intersection of two planes is not always as simple as computing the intersection of two lines. In this article, we will explain to you how to calculate it — with multiple approaches for all possible cases! Keep reading to learn:

  • What is a plane, and what is the intersection of two planes?
  • How to find the intersection of two planes: the line in parametric form.
  • How to calculate the line of intersection of two planes in the symmetric form.
  • Examples of calculation of the line of intersection between two planes.

What is a plane in geometry?

A plane in geometry is a two-dimensional surface in a 3D space, a natural extension of the concept of line in 2D geometry. Planes are flat surfaces — their curvature is zero.

A plane is uniquely identified by a point and a normal vector. The point fixes the distance between the plane and the origin of the Cartesian space, while the normal vector (a vector perpendicular to the plane) fixes the orientation of the plane in space.

Let's meet these two elements. The point on the plane is:

P0=(px,py,pz)\footnotesize P_0=(p_x,p_y,p_z)

And the normal vector is:

n=⟨a,b,c⟩\footnotesize \boldsymbol{n} = \left\langle a,b,c \right\rangle

To describe the plane in one equation, we compute a constant for the plane, dd:

d=apx+bpy+cpz​\footnotesize d = ap_x +bp_y +cp_z ​

and then we write the equation of the plane in the Cartesian space with this neat formula:

ax+by+cz=d\footnotesize ax+by+cz=d

Above, xx, yy, and zz are free to change, but only the combinations that are points on the plane satisfy the equation.

🙋 We built a tool entirely dedicated to the equation of a plane: try our equation of a plane calculator!

Before calculating the intersection of two planes: geometry of the problem

Two planes can intersect each other (unless, of course, they are parallel). When two planes intersect, a line in space is the result. This is a logical extension of the intersection of two lines in a 2D space, a much easier concept to understand:

When two lines (single-dimensional objects) intersect, we find a point, an object with zero dimensions. We can meet this reduction in dimensionality when considering planes, too. A plane is a two-dimensional object, so we expect the intersection of two planes to be a single-dimensional object, which is a line. This is indeed what we find!

🙋 You can learn about intersecting lines at our intersection of two lines calculator — and if your lines are parallel, perhaps you should visit our parallel lines calculator!

As we said before: when two planes are parallel, we can't find an intersection, and they are either separated by a given distance or totally coincide. Respectively, these situations arise if the normal vectors match (hence the planes have the same orientation) and if d1=d2d_1=d_2 (thus fixing the distance of the two planes) as well.

Finding the line of intersection of two planes involves dealing with lines in the 3D space: this is not a straightforward task! We can find the line of intersection of two planes in two ways; let's meet the first one.

How to calculate the line of intersection of two planes?

In this section, we will calculate the line of intersection of two planes as a parametric equation. This method follows a simple sequence of steps that don't depend on the coefficients of the planes' equations.

The result is in the following form:

r=(k1n1+k2n2)+λ(n1×n2)\footnotesize \boldsymbol{r} = (k_1\boldsymbol{n}_1+k_2\boldsymbol{n}_2)+ \lambda(\boldsymbol{n}_1\times\boldsymbol{n}_2)

where:

  • r=⟨x,y,z⟩\boldsymbol{r} =\left\langle x,y,z \right\rangle — Vector containing a point on the line;
  • k1k_1 and k2k_2 — Coefficients depending on the planes' equations;
  • n1\boldsymbol{n}_1 and n2\boldsymbol{n}_2 — Normalized normal vectors (i.e. ∣n1∣=∣n2∣=1|{\boldsymbol n}_1| = |{\boldsymbol n}_2| = 1);
  • n1×n2\boldsymbol{n}_1\times\boldsymbol{n}_2 — Cross product of the normal vectors; and
  • λ\lambda — A parameter that can vary freely;

The coefficients k1k_1 and k2k_2 are:

k1=d1−d2(n1⋅n2)1−(n1⋅n2)k2=d2−d1(n1⋅n2)1−(n1⋅n2)\footnotesize k_1 = \frac{d_1-d_2(\boldsymbol{n}_1\cdot\boldsymbol{n}_2)}{1-(\boldsymbol{n}_1\cdot\boldsymbol{n}_2)} \\[1em] k_2 = \frac{d_2-d_1(\boldsymbol{n}_1\cdot\boldsymbol{n}_2)}{1-(\boldsymbol{n}_1\cdot\boldsymbol{n}_2)}

By changing the value of parameter λ\lambda, we can generate every possible r\boldsymbol{r}, and thereby fully describe the line of intersection of our two planes.

How to find the intersection of two planes using the symmetric form of the line equation

We can also find the intersection of two planes in symmetric form. This result may be more familiar than the parametric form. In the symmetric form, we equate xx to the other two spatial coordinates without needing an additional parameter like λ\lambda.

To do so, we perform a series of substitutions and rearrangements of the planes' equations. In this case, there is no one-size-fits-all list of instructions. We'd better illustrate this process with an example.

Take these two planes:

  • 2x−4z=−12x - 4z = -1; and
  • x−2y+z=2x-2y+z = 2.

As you can see, in the first equation, there is no coefficient for yy: this means that the coordinates of points on this plane don't depend on yy. To find the equation of the line, follow these steps:

  1. Isolate the variable zz in the first equation: z=12x+14z = \frac{1}{2}x + \frac{1}{4}.

  2. Substitute the equation found in the previous step to zz in the second equation: x−2y+(12x+14)=2x - 2y + (\frac{1}{2}x + \frac{1}{4}) = 2.

  3. Compute the sums: (1−12)x−2y=2−14(1-\frac{1}{2})x -2y = 2 - \frac{1}{4}, i.e., 12x−2y=74\frac{1}{2} x - 2y = \frac{7}{4}.

  4. Isolate xx in the equation above to obtain x=4y+72x = 4y +\frac{7}{2}: this is how we find the line of intersection of the two planes — or even better, its projection in the xy plane.

  5. To find the projection of the intersection line in the xz plane, we can simply use the first plane equation: x=2z−12x = 2z -\frac{1}{2}.

The calculated intersection of the two planes in symmetric form is then:

x=4y+72=2z−12\footnotesize x = 4y +\frac{7}{2} = 2z -\frac{1}{2}

Only the x,y,zx, y, z combinations that define points on the line will simultaneously satisfy these two equations!

How to find the line of intersection of two planes: an example

Let's try the procedure to find the parametric equation of a line with a practical example. Take these planes:

  1. −2x+3y+4z=−1-2x + 3y +4z = -1; and
  2. 2x−y−3z=22x -y -3z = 2.

Let's find the two normal vectors:

  1. For the first plane, we have n1=⟨−2,3,4⟩\boldsymbol{n}_1 = \left\langle -2,3,4\right\rangle; and
  2. For the second plane, we have n2=⟨2,−1,−3⟩\boldsymbol{n}_2 = \left\langle 2,-1,-3\right\rangle.

We normalize the vectors by dividing each component by the norm of the vector:

∣n1∣=(−2)2+32+42=4+9+16=29\footnotesize \begin{split} |\boldsymbol{n}_1|&=\sqrt{(-2)^2+3^2+4^2}\\ &=\sqrt{4+9+16} = \sqrt{29} \end{split}

We have:

n1=⟨−229,329,429⟩\footnotesize \boldsymbol{n}_1 = \left\langle-\frac{2}{\sqrt{29}},\frac{3}{\sqrt{29}},\frac{4}{\sqrt{29}} \right \rangle

We find the norm for the second vector too:

∣n2∣=22+(−1)2+(−3)2=4+1+9=14\footnotesize \begin{split} |\boldsymbol{n}_2|&=\sqrt{2^2+(-1)^2+(-3)^2}\\ &=\sqrt{4+1+9} = \sqrt{14} \end{split}

And the normalized vector is:

n2=⟨214,−114,−314⟩\footnotesize \boldsymbol{n}_2 = \left\langle \frac{2}{\sqrt{14}},-\frac{1}{\sqrt{14}},-\frac{3}{\sqrt{14}}\right \rangle

We then calculate the coefficients we need to use in the equation for the line. For the first coefficient, we have the following:

k1=−1−2(n1⋅n2)1−(n1⋅n2)\footnotesize k_1 = \frac{-1-2(\boldsymbol{n}_1\cdot\boldsymbol{n}_2)}{1-(\boldsymbol{n}_1\cdot\boldsymbol{n}_2)}

We compute the dot product of the two normal vectors — if you need a hand with this operation, visit our dot product calculator:

n1⋅n2=⟨−229,329,429⟩⋅⟨214,−114,−314⟩=−4406−3406−12406=−19406≃−0.943\footnotesize \begin{split} \boldsymbol{n}_1\cdot\boldsymbol{n}_2 &= \left\langle-\frac{2}{\sqrt{29}},\frac{3}{\sqrt{29}},\frac{4}{\sqrt{29}} \right \rangle\\[1.2em] &\cdot \left\langle \frac{2}{\sqrt{14}},-\frac{1}{\sqrt{14}},-\frac{3}{\sqrt{14}}\right \rangle\\[1.2em] & = -\frac{4}{\sqrt{406}}-\frac{3}{\sqrt{406}}-\frac{12}{\sqrt{406}}\\[1.2em] & = -\frac{19}{\sqrt{406}}\simeq -0.943 \end{split}

And find the value of the coefficient:

k1=−1−2(−0.943)1+0.943=0.456\footnotesize k_1 = \frac{-1-2(-0.943)}{1+0.943}= 0.456

And for the second coefficient, we have:

k2=2+1(−0.943)1+0.943=0.544\footnotesize k_2 = \frac{2+1(-0.943)}{1+0.943} = 0.544

The last thing we need to find is the cross product between the normalized normal vectors. We will skip the steps here and give you only the final result — the cross product calculator will clear any doubt you can have.

n1×n2=⟨−5406,2406,−4406⟩=⟨−0.2481,0.0993,−0.1985⟩\footnotesize \begin{split} \boldsymbol{n}_1\!\times\!\boldsymbol{n}_2\!&=\!\left\langle -\frac{5}{\sqrt{406}},\frac{2}{\sqrt{406}},-\frac{4}{\sqrt{406}} \right\rangle \\[1em] &=\!\left\langle-0.2481,0.0993,-0.1985 \right\rangle \end{split}

Let's compute the products of the coefficients by the respective normalized vectors:

k1n1=0.456⟨−229,329,429⟩=⟨−0.1693,0.254,0.3387⟩\footnotesize \begin{split}k_1\boldsymbol{n}_1 &= 0.456\left\langle-\frac{2}{\sqrt{29}},\frac{3}{\sqrt{29}},\frac{4}{\sqrt{29}} \right \rangle\\[1em] &=\left\langle-0.1693,0.254,0.3387\right \rangle \end{split}

And:

k2n2=0.544⟨214,−114,−314⟩=⟨0.291,−0.1454,−0.4362⟩\footnotesize \begin{split} k_2\boldsymbol{n}_2 &= 0.544 \left\langle \frac{2}{\sqrt{14}},-\frac{1}{\sqrt{14}},-\frac{3}{\sqrt{14}}\right \rangle\\[1em] &=\left\langle 0.291,-0.1454,-0.4362 \right\rangle \end{split}

We compute the sum of these two results:

k1n1+k2n2=⟨−0.1693,0.254,0.3387⟩+⟨0.291,−0.1454,0.4362⟩=⟨0.1215,0.1086,−0.0975⟩\footnotesize \begin{split} & k_1\boldsymbol{n}_1+k_2\boldsymbol{n}_2 \\ &= \left\langle-0.1693,0.254,0.3387\right \rangle\\ &\quad+ \left\langle 0.291,-0.1454,0.4362 \right\rangle\\[.5em] &=\left\langle 0.1215,0.1086,-0.0975\right\rangle \end{split}

We can combine all these results, finally finding the line of intersection of the two planes in the parametric form:

r=(k1n1+k2n2)+λ(n1×n2)=⟨0.1215,0.1086,−0.0975⟩+λ⟨−0.2481,0.0993,−0.1985⟩\footnotesize \begin{split} \boldsymbol{r}& = (k_1\boldsymbol{n}_1+k_2\boldsymbol{n}_2)+\lambda(\boldsymbol{n}_1\times\boldsymbol{n}_2)\\[.5em] &=\left\langle 0.1215,0.1086,-0.0975\right\rangle \\ &\quad+ \lambda\left\langle-0.2481,0.0993,-0.1985 \right\rangle \end{split}

Ok, this expression is objectively clumsy; let's split it up so that we can see the individual equations cooperating for the definition of this line:

x=0.1215−0.2481λy=0.1086+0.0993λz=−0.0975−0.1985λ\footnotesize \begin{split} x &= 0.1215-0.2481\lambda\\ y&=0.1086+0.0993\lambda\\ z&=-0.0975-0.1985\lambda \end{split}

We vary λ\lambda and find combinations of points on the line of intersection.