# Thread: Equation of a Plane

1. ## Equation of a Plane

Three points are given: P = (2,1,0) , Q = (1,0,1) , R = (3,2,-1).
I'm supposed to find the plane that goes through P, Q, R in the form Ax + By + Cz = D.

My attempt:

M : (x,y,z) = P + t*PQ + s*PR , s,t R.

PQ
= Q - P = (-1,-1,1) , PR = R - P = (1,1,-1).

This gives me a system of equaton where I use the elimination method to find out what the coefficients A, B, C and D are equal to, but I'm stuck at the second step:
x = 2 - t + s
y = 1 - t + s
z = + t - s

<=>

x = 2 - t + s
-x + y = - 1
x + z = 2

Did I do something wrong? / How do I proceed from here?

2. ## Re: Equation of a Plane

Originally Posted by Cinnaman
Three points are given: P = (2,1,0) , Q = (1,0,1) , R = (3,2,-1). I'm supposed to find the plane that goes through P, Q, R in the form Ax + By + Cz = D.
Here a different way.
Let $N = \overrightarrow {PQ} \times \overrightarrow {PR}$ then the plane is $N\cdot(-P)=0$

3. ## Re: Equation of a Plane

We haven't reached cross product yet, I need to solve it using the method above.

I just don't know what happens when both t and s disappear in the third equation at step 2 of the elimination process.

4. ## Re: Equation of a Plane

So my lecturer just sent out an email saying there was an error in the question itself, now it makes sense -_-