Equation of planes

**darren86** · December 17th 2012, 06:57 AM

Hi, can anyone explain to me why the equation for a plane3 which contains the intersection line of 2 other planes(plane1 and plane2), is the sum of the equations of plane 1 and plane 2?

Note the planes all go through the origin.

ie, if plane 1 is ax+by+cz=0 and if plane 2 is dx+ey+fz=0

any plane containing the intersecting lines of plane 1 and plane 2 is:

(a+d)x + (b+e)y + (c+f)z = 0

Why?

**Plato** · December 17th 2012, 09:40 AM

Originally Posted by darren86

Hi, can anyone explain to me why the equation for a plane3 which contains the intersection line of 2 other planes(plane1 and plane2), is the sum of the equations of plane 1 and plane 2?

Notation: $\mathcal{R}=<x,y,z>$ and $P$ is a point.

Here are two planes $N\cdot(\mathcal{R}-P)=0~\&~M\cdot(\mathcal{R}-P)=0,~N\not \parallel M,$ both containing $P$ . The normals are $N~\&~M$

The equation of the line of their intersection is $P+t(N\times M)$ .

Because $(N+M)\cdot(N\times M)=0$ means the plane $(N+M)\cdot(\mathcal{R}-P)=0$ contains the line of intersection and has normal $N+M$ .

NOTE that answer is not unique. But it explains your question.

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