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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

enter image description here

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 ?

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    $\begingroup$ 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 $\alpha^2+\beta^2 >0$ $\endgroup$ – Jaroslaw Matlak Jul 23 '18 at 15:58
  • $\begingroup$ @JaroslawMatlak but how do I derive it ?. could u pls guide how do I bring in those parameters $\alpha$ and $\beta$ ? $\endgroup$ – ss1729 Jul 23 '18 at 16:06
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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 $\pi_2$).

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  • $\begingroup$ 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. $\endgroup$ – David K Jul 24 '18 at 12:05
  • $\begingroup$ 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. $\endgroup$ – ss1729 Jul 24 '18 at 12:47
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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 $2$).

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

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  • $\begingroup$ thnx. but why do we multiply with $\lambda$ how can we explain that ? $\endgroup$ – ss1729 Jul 23 '18 at 17:43
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    $\begingroup$ @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. $\endgroup$ – poyea Jul 24 '18 at 6:10
  • $\begingroup$ ok thanx. so is my explanation be taken as well defined?. pls check i have edited OP. $\endgroup$ – ss1729 Jul 24 '18 at 7:00
  • $\begingroup$ 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. $\endgroup$ – poyea Jul 24 '18 at 8:32
  • $\begingroup$ 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. $\endgroup$ – ss1729 Jul 24 '18 at 9:09
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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.

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