# Plane through the intersection of two given planes.

If I have two planes $$\mathbf r\cdot\mathbf {\hat n_1}=p_1\\ \mathbf r\cdot\mathbf {\hat n_2}=p_2$$

If they intersect somewhere then the intersection will form a line, if $$\mathbf r'$$ is the position vector of a point on the intersection line then I have $$\mathbf r'\cdot\mathbf {\hat n_1}=p_1\\ \mathbf r'\cdot\mathbf {\hat n_2}=p_2$$

If I combine these two equations I would get $$\mathbf r'\cdot(\mathbf {\hat n_1}+\mathbf{\hat n_2})=p_1+p_2$$ This looks like a vector equation of a plane.

But if we see in internet or in books we would find there is scalar $$\lambda$$ multiplied.

$$\mathbf r'\cdot(\mathbf {\hat n_1}+\lambda\mathbf{\hat n_2})=p_1+\lambda p_2$$

So my question is why there is $$\lambda$$ there? Please also give me the geometrical image of the plane which is passing through the intersection of those two planes. I'm not able to imagine how would that plane look like.

• Your notation is difficult to read. The intersection of two planes can also be the entire plane. Let's assume that is not the case. To get the vector corresponding to the line of intersection you just have to take the cross product of the normal vectors to each plane. Mar 22, 2019 at 14:55
• @JohnDouma I'm sorry for my bad Latex. Mar 22, 2019 at 15:00
• Mar 22, 2019 at 15:01

We have to make some considerations:

1. The intersection of two planes could be empty if the planes are parallel, but you already state that you assume an intersection, so this is not a problem.
2. The intersection of two planes which are the same is just the plane itself. We will deal with this case later.
3. Suppose now you have two distinct, non-parallel planes. You write the equations of each plane as $$\vec{r}\cdot\hat{n}_1=p_1$$ and $$\vec{r}\cdot\hat{n}_2=p_2$$. Now, if I multiply each of these equations by a constant, the equations remain true. For instance, $$\vec{r}\cdot\hat{n}_1=p_1$$ implies $$\vec{r}\cdot(A\hat{n}_1)=Ap_1$$, and similarly I can get $$\vec{r}\cdot(B\hat{n}_2)=Bp_2$$.
4. These two planes are distinct and non-parallel, so they intersect in a line. As you say, points on this line have to satisfy both plane equations simultaneously, so I can describe the line by the system of equations $$\begin{cases} \vec{r}\cdot(A\hat{n}_1)=Ap_1 \\\vec{r}\cdot(B\hat{n}_2)=Bp_2 \end{cases}.$$
5. As you also pointed out, the combination of these equations $$\vec{r}\cdot(A\hat{n}_1+B\hat{n}_2)=Ap_1+Bp_2$$ looks like the equation of a plane for given $$A$$ and $$B$$, because it is so. This is not the equation of a line, many more points satisfy it. Let us divide through by $$A$$ (assuming it is not zero), and denote $$B/A$$ by $$\lambda$$. We obtain: $$\vec{r}\cdot(\hat{n}_1+\lambda\hat{n}_2)=p_1+\lambda p_2.$$ Once again, for a fixed $$\lambda$$, this is the equation of a plane.
6. However, we obtain a line if we require that this equation is true for any value of $$\lambda$$. Remember that the values of $$A$$ and $$B$$ were arbitrary, and therefore so is $$\lambda$$. The intersecting line should satisfy this new equation for all $$\lambda$$. For instance, if I choose $$\lambda=0$$ and $$\lambda=1$$, I recover the system: $$\begin{cases} \vec{r}\cdot\hat{n}_1=p_1 \\\vec{r}\cdot(\hat{n}_1+\hat{n}_2)=p_1+p_2 \end{cases},$$ which is equivalent to the system we had in point 4.
7. This is all still true if the two planes are the same, but in this case the system of equations is redundant because $$\hat{n}_1=\hat{n}_2$$ and $$p_1=p_2$$. This gives $$\begin{cases} \vec{r}\cdot(A\hat{n}_1)=Ap_1 \\\vec{r}\cdot(B\hat{n}_1)=Bp_1 \end{cases}$$ and $$\vec{r}\cdot[(\lambda+1)\hat{n}_1]=(\lambda+1) p_1,$$ which only define the same plane again, not a line.
8. As for the picture, I found this image online: The red and green planes intersect in a line, and the blue one is another plane which passes through the line.

I hope this helps!

• Your answer is very clear and helpful. Thanks for being so kind. Mar 22, 2019 at 16:06
• You're very welcome.
– R_B
Mar 22, 2019 at 16:09
• @R_B in point 5 how do we know that it is not the equation of a line and that many more points satisfy it? May 16, 2019 at 11:11
• @SuzieWaters look at the second equation in point 5, $\vec{r}\cdot(\hat{n}_1+\lambda\hat{n}_2)=p_1+\lambda p_2$. I obtained this equation from the first one through division by a constant, so the equations are equivalent. Now I can define new variables to get $\vec{r}\cdot\hat{n}_3=p_3$, which is the usual vector equation of a plane.
– R_B
May 16, 2019 at 11:18
• @HarryHolmes It's easier to think in 2D, with the equations of two lines intersecting a single point. If you equate both line equations, you get another line equation. To find the point, you need to solve a system of two equations, not just one equation. In 3D is the same, a single equation defines only a plane, no matter how you get it. You can also think about restrictions: in 3D, one linear equation gives you a way to find one of the coordinates from the other two. If you can choose two coordinates freely, and only the third one is fixed, you have a plane.
– R_B
Apr 23, 2021 at 10:02

$$\mathbb r\mathbb n=p$$ and $$\mathbb r(\lambda\mathbb n)=\lambda p$$ denote the same plane. Also any combination

$$\mathbb r(\lambda\mathbb n_1+\mu\mathbb n_2)=\lambda p_1+\mu p_2$$ is the equation of a plane.

Now if we consider a point $$\mathbb r$$ that satisfies the equation of two planes,

$$\mathbb r\mathbb n_1=p_1$$ and $$\mathbb r\mathbb n_2=p_2,$$ then

$$\mathbb r(\lambda\mathbb n_1+\mu\mathbb n_2)=\lambda\mathbb r\mathbb n_1+\mu\mathbb r\mathbb n_2=\lambda p_1+\mu p_2$$ is true.

So any point common to the two planes belongs to the combined plane.

Imagine a plane that rotates around the intersection line.

Lambda is used for direction here We can see the plane passes through there intersection can be perpendicular to them or at 120 degrees, so its purpose is to give different planes for different values of lambda.