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!
Best Answer
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: