Systems of Linear Equations

  • Thread starter Nusc
  • Start date
  • #1
753
2
The question asks: Find the equation of the plane that passes throug the points (1,2,3), (3,-1,3), and (5,0,7). (Hint: Recall that the general equation of a plane is ax + by +cz =d.)

You could make 2 vectors with those three points then find the normal by taking the cross product of those two vectors then plug them into the general equation with xo being the ___ (one point that makes them both vectors - what's the word?)

But say you had no previous knowledge of vector geometry, how would one going about doing this problem?
 
Last edited:

Answers and Replies

  • #2
cronxeh
Gold Member
961
10
Well I suppose you'd need one more point to construct the 0 degrees of freedom linear system..
so far you got
a + 2b + 3c = d
3a -b + 3c = d
5a +0b + 7c = d
 
  • #3
AKG
Science Advisor
Homework Helper
2,565
4
If ax + by + cz = d is the equation of the plane, then those points are solutions to the equation. You can basically set up 3 equations:

1a + 2b + 3c - 1d = 0
3a - 1b + 3c - 1d = 0
5a + 0b + 7c - 1d = 0

Row reduce the corresponding matrix and you'll solve for a, b, c, and d all in terms of d (or one of the parameters). And of course, there won't be a unique solution, the solution will indeed be in terms of d since

x + y + z = 1

and

2x + 2y + 2z = 2

describe the same plane.
 
  • #4
753
2
Well the answer is 3x + 2y - 2z = 1, which I obtained using methods of vector geometry but when I solve the linear system I'm left with
[tex]\mbox {~}\left[\begin{array}{ccc|r}1&0&0&3\\0&1&0&2\\0&0&1&-2\end{array}\right][/tex]

How do you get d?
 
Last edited:
  • #5
AKG
Science Advisor
Homework Helper
2,565
4
You should have a (3 x 4)-matrix with a (3 x 1)-matrix of zeroes augmented on the end. Note that you have 3 equations and 4 unknowns.
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,833
956
Nusc said:
Well the answer is 3x + 2y - 2z = 1, which I obtained using methods of vector geometry but when I solve the linear system I'm left with
[tex]\mbox {~}\left[\begin{array}{ccc|r}1&0&0&3\\0&1&0&2\\0&0&1&-2\end{array}\right][/tex]

How do you get d?
You don't get d! As you were told before, the form ax+by+cz= d is not unique. You can divide or multiply the entire equation by any number (other than 0) to get a different equation for the same plane. In particular, as long as d is not 0 you can divide through by it to get an equation of the form Ax+ By+ Cz= 1 (A= a/d, B= b/d, C= c/d). If start by assuming that form, you get your answer.
 

Related Threads on Systems of Linear Equations

  • Last Post
Replies
5
Views
851
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
1K
Replies
2
Views
4K
Replies
2
Views
2K
Replies
4
Views
2K
Replies
1
Views
1K
Replies
7
Views
4K
Replies
8
Views
7K
Replies
5
Views
3K
Top