Equation perpendicular to 2 non-parallel planes

Question

Good day sirs! Can you help me with this questions?

Find the general equation of the plane:

(1) Through $(3,0,-1)$ and perpendicular to each of the planes $x-2y+z=0$ and $x+2y-3z-4=0$

(2) Perpendicular to the plane $x+2y-3z-7=0$ and containing the points $(2, 1, 4)$ and $(1,1,0)$

I tried to answer this, yet I believe there is no such plane perpendicular to two non-parallel planes. (Am I wrong? :) )

My professor tried to explain how to get the answer to my classmates and they all seem to agree and "enlightened", but I cannot picture out in my mind how is it possible to have a plane with such conditions, so I'm still skeptic about it.

*Can you include graphs also? :)

Answer 1

In the first part you need the cross product of the two plane normals to find the required normal to the plane, so that you can write down the equation

In the second part, find the vector joining the two points and do the cross product of this and the normal to the given plane,cso you can write down the equation in a similar way as before.

Answer 2

For part #1:

Let's call the two given planes $A$ and $B$, and the desired plane $C$.

If $C$ is perpendicular $A$ and $B$, then the normal $N_C$ of $C$ is perpendicular to the normals $N_A$ and $N_B$ of $A$ and $B$. Draw a picture to convince yourself that this is true.

So $N_C$ is parallel to the cross product $N_A \times N_B$.

In your case, $A$ is the plane $x-2y+z=0$, so $N_A = (1,-2,1)$, and $B$ is the plane $x+2y-3z-4$, so $N_B = (1,2,-3)$.

Then $N_C = N_A \times N_B = (4,4,4)$.

The plane perpendicular to $(4,4,4)$ through the point $(3,0,-1)$ is $(x-3,y,z+1)\cdot (4,4,4) =0$.

Answer 3

I think the following picture can explain your doubt.