Thread: Finding the equation of a plane given three points.

1. Finding the equation of a plane given three points.

So the points given are A = (5,5,3), B = (2,8,9), and C = (1,0,5) and I have to find the equation of the plane.

I know that in order to do this, you have to find the line that's perpendicular to these vectors and to do that, you can take the cross product of two of them. I decided to use AB and AC.

So here's what I have:

AB = -3i+3j+6k
AC = -4i-5j+2k

AB x AC = 36i+18j+27k -> 36x+18y+27z = 351.

But when you plug the other points into this equation, I don't get the same answer, which I should. I must've gone wrong somewhere, but I'm not sure what I did wrong. It could be a simple arithmetic thing, but either way, I'm unsure. Could someone help me out with this problem and/or tell me what I did wrong? Any help would be appreciated!

2. Originally Posted by Rumor
So the points given are A = (5,5,3), B = (2,8,9), and C = (1,0,5) and I have to find the equation of the plane.

I know that in order to do this, you have to find the line that's perpendicular to these vectors and to do that, you can take the cross product of two of them. I decided to use AB and AC.

So here's what I have:

AB = -3i+3j+6k
AC = -4i-5j+2k

AB x AC = 36i-18j+27k -> 36x-18y+27z = 171.
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