# Finding cartesian equation of plane

1. Jun 17, 2014

### thatoneguy6531

1. The problem statement, all variables and given/known data
determine the Cartesian equation of the plane through the points (3,0,1) and (0,1,-1) and perpendicular to the plane with equation x-y-z+1 = 0

2. Relevant equations

3. The attempt at a solution
Well I know the normal of the plane (a,b,c) dotted with (1,-1,-1) = 0 which gives me a-b-c=0
but other than that Im stuck

2. Jun 17, 2014

### Simon Bridge

You need to write out the description in terms of the maths.
You know the standard form for the cartesian equation of the plane (or the vetor form - whichever you are most comfortable with). So write it out.

That equation has to satisfy some conditions - write out what those mean.

i.e. point (3,0,1) is on the plane - so if you put x=3, y=0, z=1 into the equation, how does it come out?

The rest is simultaneous equations.

3. Jun 18, 2014

### HallsofIvy

Staff Emeritus
(a, b, c).(0, 1,-1)= b- c= 0 so you know b= c. The equation must be of the form a(x- 3)+ by+ b(z- 1)= 0 and you know that x= 0, y= 1, z= -1 must satisfy that: -3a+ b- 2b= 0. Of course, any multiple of (a, b, c) will give the same plane so you can take either a or b to be whatever you like.

4. Jun 18, 2014

### thatoneguy6531

Ok, thanks

5. Jun 18, 2014

### Simon Bridge

Lucky person you - HoI does not normally do people's homework for them...

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