Find Equation of a Third Plane Which Interacts With Two Given Planes

Question

I have an online course asking me to find an equation of a third plane which intersects with two given planes (P1: x + y + 3z − 2 = 0 and P2: x − y + 2z = 0) in one point, and in one line (so two different answers).

We have covered nothing like this in the content provided and I can't find anything like this on the internet (including the database on mathematics stackexchange)

I'm not asking for straight up answers, but could someone please show the path or show me where I am supposed to start?

What I tried: I thought that if all the planes are perpendicular, then the planes would meet at one point, thus I found the cross product of the two normals (1,-1,3) x (1,-1,2) and found (5,1,-2) to be the normal, thus I created this into a plane P3: x + y = 0, but thought that maybe the D value (Ax + Bx + Cy + D = 0) is supposed to equal something, thus I abandoned such solution.

Also for the intersection at one line, I believe I could just make the third plane perpendicular to one of the planes and that would create a system where the planes intersect at a line. Is this right?

Here's the actual question: Determine the equation of a plane, P3, that intersects the planes P1: x + y + 3z − 2 = 0 and P2: x − y + 2z = 0 in a point; b line.

Please help!

Thanks!

Answer 1

If a plane is given in the so called point-normal form:

$$ax+by+cz=d$$ then the vector

$$\vec n=\begin{bmatrix} a\\ b\\ c \end{bmatrix}$$

is perpendicular to that plane.

We have two planes and the normal vectors are

$$\vec n_1=\begin{bmatrix} 1\\ 1\\ 3\end{bmatrix}\text{ and } \ \vec n_2=\begin{bmatrix} 1\\ -1\\ 2 \end{bmatrix}.$$

Taking the vector product of these two vectors we get a new vector which is perpendicular to both of the normal vectors and it is parallel to both of the planes; and it is then parallel to their intersection line. So,

$$\vec d=\vec n_1\times \vec n_2=\begin{vmatrix} \vec i&\vec j&\vec k\\ 1&1&3\\ 1&-1&2 \end{vmatrix}=\begin{bmatrix} 5\\ 1\\ -2\end{bmatrix}$$

We have the direction vector of the intersection line. Any vector perpendicular to $\vec d$ could serve as a normal vector of a plane that contains the intersection line of the two planes. $\vec n_1$ and $\vec n_2$ would be such vectors but we don't want to use them (or any scalar multiples of them) because we want a plane different from the planes given. In order to find a vector perpendicular to $\vec d$ we need to solve the following scalar product equation

$$\vec d\cdot \vec n_3=\vec d\cdot\begin{bmatrix} x_{n_3}\\ y_{n_3}\\ z_{n_3} \end{bmatrix}=5x_{n_3}+y_{n_3}-2z_{n_3}=0.$$

We may freely choose, say, $x_{n_3}=0$ and $y_{n_3}=2$. Then $z_{n_3}=1$ is the right choice:

$$\vec n_3=\begin{bmatrix} 0\\ 2\\ 1\end{bmatrix}$$

will be perpendicular to $\vec d$.

In order to get the equation of the third plane we need a point on the intersection line. For example, a point belonging to $t=0$, a common point of the two planes given satisfies that requirement. So the normal vector is $\vec n_3$ and a point on the third plane is $(1,1,0).$

Let $(x,y,z)$ be an arbitrary point of the third plane. Then the vector

\begin{bmatrix} x-1\\ y-1\\ z \end{bmatrix}

will be parallel to it and the scalar product of this vector and $n_3$ will be zero:

$$\vec d \cdot \begin{bmatrix} x-1\\ y-1\\ z \end{bmatrix}= \begin{bmatrix} 0\\ 2\\ 1\end{bmatrix}\cdot \begin{bmatrix} x-1\\ y-1\\ z \end{bmatrix}=0.$$

Hence, the point-normal equation of a suitable third plane is

$$2(y-1)+z=2y+z-2=0.$$

Let's check if the intersection line of the first two plane is in the third one. The equation of the intersection line is

$$\begin{bmatrix} x(t)\\ y(t)\\ z(t)\end{bmatrix}=\begin{bmatrix} 5\\ 1\\ -2\end{bmatrix}t+\begin{bmatrix} 1\\ 1\\ 0\end{bmatrix}.$$

Substituting $x(t)$, $y(t)$, and $z(t)$ into the equation of the third plane will result in $0$. Finally we can see that $n_3$ is not a scalar multiple of either $n_1$ or $n_2$.

The following figure illustrates what we have been doing:

Answer 2

For a plane that intersects two other planes at a line in $3$ dimensional Euclidean space:

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Suppose $n_1$ is a normal for plane $P_1$ that lies in $3$ dimensional Euclidean space and $n_2$ is a normal for $P_2$. Then $n_1 \times n_2$ is normal to the normals of both planes. It is normal to $n_1$, but what does that mean?

If it's normal to the normal of the plane $P_1$, it must be a vector that is tangent to $P_1$.

So $n_1 \times n_2$ is tangent to both planes, and so it is a direction vector for the line that intersects both planes, if such a line exists.

So we can find the direction vector to the line that two planes intersect by taking the cross product of their normals.

To find a point on the line is easy. Take $z=0$ or set anything else to a constant, and you should get a solvable system of equations the other coordinates of a point that lies on the line of intersection of the planes.

Once you have this information, the line of intersection is,

$$\vec r(t)=\vec v t+\vec w$$

Where $\vec v=\vec n_1 \times \vec n_2$ and $\vec w$ is the position vector that corresponds to any specific point you found that is on the line of intersection.

To find the equation of any plane that contains this line (but not the two known planes) pick any other point $A$ not on the line nor on either already known planes. Then pick a point on the line and call this $B$. We will calculate the equation of the plane that contains the line and the new point. $\vec v \times \vec BA$ is a normal to such a plane. Then the equation is,

$$(\vec v \times \vec BA) \cdot (\langle x,y,z \rangle-\vec B)=0$$

For a plane in $3$ dimensional Euclidian space that intersects two other planes at a unique point.

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In my opinion, It's better to think about this problem in terms of a system of equations,

Two planes can be,

$$ax+by+cz=d$$

$$ex+fy+gz=h$$

Let's assume you already know these planes. Another plane can be,

$$i x+j y+k z=l$$

In order for these three to intersect at one unique point there must be a unique solution to this system of equations. Which means the matrix for this system of equations has to be invertible, hence implying the determinant of coefficients is nonzero. That gives some restriction on $i, j, k$. Then, noting that restriction pick any $i, j,k$ that does not conflict with the restriction and pick any $x,y,z$ that lies on the line of intersection that was found in the previous parts. Substitute those into the third equation, that will give you $l=i x+j y+k z$. So now you know $i,j,k,l$ that will work and have $3$ planes with a unique solution.