# Systems of Linear Equations

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?

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cronxeh
Gold Member
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

AKG
Homework Helper
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.

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
$$\mbox {~}\left[\begin{array}{ccc|r}1&0&0&3\\0&1&0&2\\0&0&1&-2\end{array}\right]$$

How do you get d?

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AKG
Homework Helper
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.

HallsofIvy
$$\mbox {~}\left[\begin{array}{ccc|r}1&0&0&3\\0&1&0&2\\0&0&1&-2\end{array}\right]$$