Creating a perpendicular plane

[samako]

I've been having trouble with lots of basic geometry, including a lot of things regarding planes and intersection of planes.

My question is, how would I get the equation of a plane (given a point on the plane), which is perpendicular to the line of intersection of two other planes?

I am not only interested in the answer, but more into the reasoning and things... because I do get quite a few questions like these, and I think that it would be better for me to actually understand what I'm working with. So the more detailed, and informative explanation (with perpendicular and parallel planes), the better.

 

[TD]

Hint: a plane \alpha :ax + by + cz + d = 0 has a normal vector which is perpendicular to the plane given by (a,b,c).

 

[marlon]

I've been having trouble with lots of basic geometry, including a lot of things regarding planes and intersection of planes.
My question is, how would I get the equation of a plane (given a point on the plane), which is perpendicular to the line of intersection of two other planes?
I am not only interested in the answer, but more into the reasoning and things... because I do get quite a few questions like these, and I think that it would be better for me to actually understand what I'm working with. So the more detailed, and informative explanation (with perpendicular and parallel planes), the better.

Take two points of the intersecting line and subtract their coordinates. You get a new vector (a,b,c) that denotes the direction of that line. Use the new vector as the normal vector in the plane equation, just like TD pointed out. There is still one unkown (d) in \alpha :ax + by + cz + d = 0. You can find this d by plugging in the given coordinates of a point that belongs to the plane you are looking for

marlon