HelpGiven a common intersection point, create 3 different planes

[silentfire]

Homework Statement

Given a common intersection point (3,4,5), find 3 different planes.

Homework Equations

None

The Attempt at a Solution

What I did is let

a₁x+a₂y+a₃z=a
b₁x+b₂y+b₃z=b
c₁x+c₂y+c₃z=c

3=D_x/D 4=D_y/D 5=D_z/D

I set D=2, therefore D_x=6 D_y=8 D_z=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.

 

[Dustinsfl]

If you want 3 different planes that intersect at (3,4,5), why not just use z=5, y=4, and x=3?

 

[silentfire]

Thanks a lot dustin for your reply.

If i want to justify that this is indeed the intersection point of the 3 different planes you mentioned, can I apply Cramer's rule and then do the dot and cross products with the normal vectors of the 3 planes. If the determinant is indeed not equal to 0, then the intersection point is justified?

 

[Dustinsfl]

Well if we use those equations then are we would have x+0y+0z=3; 0x+y+0z=4; 0x+0y+z=5. Which would make our determinant 1.