# Derive Equation of Plane passing through the intersection of two planes

How to derive the equation of the plane passing through the intersection of two given planes.

Lets say we have given two planes $$\pi_1 : \vec{r}.\hat{n}_1=d_1\\ \pi_2 : \vec{r}.\hat{n}_2=d_2\\$$ where $\hat{n}_1, \hat{n}_2$ : unit vectors normal to the planes $\pi_1, \pi_2$ and $d_1, d_2$ : perpendicular distances from the origin.

The position vector of any point on the line of intersection must satisfy both the equations.

So far good, I understand this. But, from here how do I prove that any plane passing through the intersection of the planes is:

$$\boxed{ \pi_3 : \vec{r}.(\hat{n}_1+\lambda \hat{n}_2)=d_1+\lambda d_2\\ \qquad\qquad\quad\text{OR}\\ \pi_3 : \vec{r}.(\alpha\hat{n}_1+\beta \hat{n}_2)=\alpha d_1+\beta d_2 }$$

My Understanding

From the figure

and parallelogram law of vector addition, the normal of the plane passing through the intersection of $\pi_1$ and $\pi_2$ will be some linear combination of $n_1$ and $n_2$. ie, $\hat{n}_3=\alpha\hat{n}_1+\beta\hat{n}_2$. And any point should satisfy equations of $\pi_1$ and $\pi_2$. Thus, $$\vec{r}.\hat{n}_3=D \implies \vec{r}.(\alpha\hat{n}_1+\beta\hat{n}_2)=\alpha\vec{r}.\hat{n}_1+\beta\vec{r}.\hat{n}_2=\alpha d_1+\beta d_2=D\\\color{red}{ \implies \vec{r}.(\alpha\hat{n}_1+\beta\hat{n}_2)=\alpha d_1+\beta d_2}$$

Is it the right explanation of the derivation ?

• Actually the plane equation should look more like: $\pi_3:r.(\alpha n_1 + \beta n_2) = \alpha d_1 + \beta d_2$, for any $\alpha, \beta$ such that $\alpha^2+\beta^2 >0$ – Jaroslaw Matlak Jul 23 '18 at 15:58
• @JaroslawMatlak but how do I derive it ?. could u pls guide how do I bring in those parameters $\alpha$ and $\beta$ ? – ss1729 Jul 23 '18 at 16:06

I think your figure gives a good intuition why the formulas work, provided that the planes $\pi_1$ and $\pi_2$ are distinct and not parallel. (If $\pi_1$ and $\pi_2$ are identical or parallel to each other, the formulas for $\pi_3$ give a plane parallel to $\pi_1$ and $\pi_2$.)

Assuming $\pi_1$ and $\pi_2$ are distinct and not parallel, take a plane $\pi_\perp$ perpendicular to the line of intersection; any plane through the line of intersection must have a normal vector parallel to that plane. The two normal vectors in the figure are both parallel to $\pi_\perp$, and they span the two-dimensional space of vectors consisting of all vectors parallel to $\pi_\perp$. In other words, the set of non-zero linear combinations of the two normals is exactly the set of normals to all planes through the line of intersection.

That gives you the formula $$\pi_3 : \vec{r}.(\alpha\hat{n}_1+\beta \hat{n}_2)=\alpha d_1+\beta d_2,$$

and the other formula can be derived from that as long as $\alpha \neq 0$ (that is, it can represent every such plane except the given plane $\pi_2$).

• I wish to note that while I like this intuition (otherwise I wouldn't have written it up), Yves Dauoust's answer makes a much more rigorous argument and better qualifies as a derivation of the formulas. – David K Jul 24 '18 at 12:05
• great. this is what i was looking for. plane through the line of intersection is the linear combination of the given two planes is little hard to make sense. I think If you think of normal vectors as u explained it'd give a better insight into the problem. – ss1729 Jul 24 '18 at 12:47

All the points that satisfy the first (second) equation belong to the first (second) plane.

You are free to form a linear combination of two equations, for instance with the coefficients $1$ and $\lambda$.

$$\vec r\vec n_1+\lambda\,\vec r\vec n_2=d_1+\lambda\,d_2,$$ which can be written

$$\vec r\,(\vec n_1+\lambda\,\vec n_2)=d_1+\lambda\,d_2.$$ The resulting equation has the shape of a plane equation. Furthermore, any point that satisfies the two given equations will also verify the combined one. So the new equation describes a plane that contains the intersection of the two given planes.

In fact, if you vary $\lambda$, you get an infinity of different planes (among which the plane $1$ for $\lambda=0$; this parameterization does not allow to include the plane $2$).

If you admit that all the possible solution planes are defined by the line of intersection and a point outside this line, then you can plug the point in the combined equation and draw the value of $\lambda$. If the point is outside the plane $2$, there is a solution. Hence the equation describes all possible planes (but $2$).

• @ss1729: this is explained in my answer. – Yves Daoust Jul 23 '18 at 17:40
• Pls check, i have edited OP. is it a better explanation for the derivation ? – ss1729 Jul 24 '18 at 7:45
• @ss1729: it is better to enforce $\alpha+\beta=1$. – Yves Daoust Jul 24 '18 at 7:46
• srry did not get ur point. Why $\alpha+\beta =1$ ? – ss1729 Jul 24 '18 at 7:48
• @ss1729: because two parameters is too much. – Yves Daoust Jul 24 '18 at 8:04

Since $\vec t$ is on the intersecting line, $\vec t$ satisfies $\pi_1$ and $\pi_2$. So it also satisfies a linear combination of $\pi_1$ and $\pi_2$. $$\begin{cases}\vec{t}\cdot \hat{n}_1=d_1\\ \vec{t}\cdot\hat{n}_2=d_2\end{cases}$$ If you multiply $\pi_2$ with a real $\lambda$, and add them together, you'll have $$\vec{t}\cdot \hat{n}_1+\lambda(\vec{t}\cdot\hat{n}_2)=d_1+\lambda d_2$$ and thus $$\vec{t}\cdot (\hat{n}_1+\lambda \hat{n}_2)=d_1+\lambda d_2.$$ You can of course do $\alpha\pi_1+\beta\pi_2$, but we want to ensure $\alpha,\beta$ are not all zero: say taking $\alpha=1,\beta=\lambda\in\mathbb{R}$.

• thnx. but why do we multiply with $\lambda$ how can we explain that ? – ss1729 Jul 23 '18 at 17:43
• @ss1729 You want to obtain a linear combination of the planes. For $\vec t\cdot(\alpha\hat{n}_1+\beta \hat{n}_2)$, you'll have an assumption that at least one of $\alpha,\beta$ is non-zero. If you try to make an analogy with vectors in 2D: plane + another plane = new plane, just as vector addition. As Yves has mentioned: this is just a way of parameterisation. – poyea Jul 24 '18 at 6:10
• ok thanx. so is my explanation be taken as well defined?. pls check i have edited OP. – ss1729 Jul 24 '18 at 7:00
• I think I'm not quite understand your question. Generally the "steps" to deal with the problem are: 1. Write $\pi_1+\lambda\pi_2$ 2. From some given conditions, you can find a particular, suitable $\lambda$. Maybe you're confused with Step 1. So if you do $\alpha\pi_1+\beta\pi_2$ you may need two given conditions. @ss1729 Also $\pi_1+\lambda\pi_2$ represents a set of planes. – poyea Jul 24 '18 at 8:32
• i am confused with step 1, ie. i'm trying to make sense to $\pi_3=\alpha\pi_1+\beta\pi_2$. Thats why i was trying to understand it from the vectors. as defined in the last section of OP. – ss1729 Jul 24 '18 at 9:09

Given two distinct intersecting planes

$$\Pi_1\to a_1x+b_1y+c_1z+d_1=0\\ \Pi_2\to a_2x+b_2y+c_2z+d_2=0$$

and now considering $\lambda_i \ne 0$, given

$$\Pi'_1\to \lambda_1\left(a_1x+b_1y+c_1z+d_1\right)=0\\ \Pi'_2\to \lambda_2\left(a_2x+b_2y+c_2z+d_2\right)=0$$

The solution set for $\Pi_1\cap\Pi_2$ and for $\Pi'_1\cap\Pi'_2$ are identical. This solution set corresponds to the intersection line.

Adding $\Pi'_1$ and $\Pi'_2$ we obtain

$$\Pi_{\lambda} \to (\lambda_1 a_1+\lambda_2 a_2)x+(\lambda_1 b_1+\lambda_2 b_2) y+(\lambda_1 c_1+\lambda_2c_2)z +\lambda_1d_1+\lambda_2 d_2 = 0$$

now $\Pi_{\lambda}\cap\Pi_1 = \Pi_{\lambda}\cap\Pi_2$ hence by construction $\Pi_{\lambda}$ contains any plane which has common intersection with $\Pi_1$ and $\Pi_2$

NOTE

making $\lambda_1 = 1$ and $\lambda_2 = \lambda$ we have the proposed situation.