Given a common intersection point (3,4,5), find 3 different planes.
The Attempt at a Solution
What I did is let
3=Dx/D 4=Dy/D 5=Dz/D
I set D=2, therefore Dx=6 Dy=8 Dz=10,
I was stuck then, no matter how I did I just got a bunch of unknown.
This question is supposed to find any 3 linear equations(aka 3 planes) that satisfy the intersection point.
Well, who doesn't know how to find the intersecting point using Cramer's rule...
P/S: This question comes up in my calculus 3 question, and I have not taken any algebra course yet, so I only know the condition for Cramer's rule for a common intersection point to take place which is D not equal to 0.