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Help!Given a common intersection point, create 3 different planes

  • Thread starter silentfire
  • Start date
  • #1

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=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.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
699
5
If you want 3 different planes that intersect at (3,4,5), why not just use z=5, y=4, and x=3?
 
  • #3
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?
 
  • #4
699
5
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.
 

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