The general expression of plane through the intersection of other two planes

Question

For two planes: $$A_{1}x+B_{1}y+C_{1}z+D_{1}=0 $$ $$A_{2}x+B_{2}y+C_{2}z+D_{2}=0$$ Prove that any plane going through the intersection line of the previous planes could be expressed like where $\lambda,\mu$ are not simultaneously zeros : $$\lambda(A_{1}x+B_{1}y+C_{1}z+D_{1})+\mu(A_{2}x+B_{2}y+C_{2}z+D_{2})=0$$

This is quite a usual knowledge, but I don't know where to start to prove this. Maybe we can use the linear algebra to solve it? Anyone knows it?

Answer 1

Hint. First you will need to assume that the given planes are not parallel.

1. Prove that your third equation is the equation of some plane.
2. Prove that any point $(x,y,z)$ which lies on the intersection of the first two planes is also on the third plane.

Answer 2

a) If the arbitrary plane is one of the two given planes, its equation will be given by setting $\lambda=0, \mu=1$ or $\lambda=1, \mu=0$.

b) Else there will be a point $P=(a,b,c)$ of the arbitrary plane not on the two given planes and you can take $$\lambda=-(A_{2}a+B_{2}b+C_{2}c+D_{2}) \neq0, \quad \mu=A_{1}a+B_{1}b+C_{1}c+D_{1} \neq0 $$

Remark
As David already correctly said, you have to start with two distinct, non-parallel planes.
The condition for that is that the non-zero vectors $(A_{1},B_{1},C_{1}), (A_{2},B_{2}, C_{2})\in \mathbb R^3$ be non-proportional.