## Finding the equation of a plane passing through 3 points

1. The problem statement, all variables and given/known data
Find the equation of the plane passing through point A(2,1,0), B(3,-1,5) and C(2,2,1)

2. Relevant equations
Um..I don't know?

3. The attempt at a solution
Vector AB=(3 -1 5)^T-(2 1 0)^T=(1 -2 5)^T
Vector AC=(2 2 1)^T-(2 1 0)^T=(0 1 1)^T
perpendicular vector n= vector ABxvector AC=(-2*1-5*1 5*0-1*1 1*1--2*0)^T=(-7 -1 1)

since n=(a b c)^T will give a plane with equation ax+by+cz=d, I had assumed the equation of the plane would therefore be -7x-y+z=d, but it turns out it's actually 7x+yx-z=15

Why are the signs inverted? (I sort of missed a few classes so I may be unaware of some fundamental concepts here)
and how do I find d (which turns out to be 15)
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Mentor
 Quote by Luscinia 1. The problem statement, all variables and given/known data Find the equation of the plane passing through point A(2,1,0), B(3,-1,5) and C(2,2,1) 2. Relevant equations Um..I don't know? 3. The attempt at a solution Vector AB=(3 -1 5)^T-(2 1 0)^T=(1 -2 5)^T Vector AC=(2 2 1)^T-(2 1 0)^T=(0 1 1)^T perpendicular vector n= vector ABxvector AC=(-2*1-5*1 5*0-1*1 1*1--2*0)^T=(-7 -1 1) since n=(a b c)^T will give a plane with equation ax+by+cz=d, I had assumed the equation of the plane would therefore be -7x-y+z=d, but it turns out it's actually 7x+yx-z=15 Why are the signs inverted? (I sort of missed a few classes so I may be unaware of some fundamental concepts here) and how do I find d (which turns out to be 15)
What do you get if you multiply both sides of 7x+y-z=15 by -1 ?
 Getting to "d": take one point and solve for d to put it in the plane. If you use a different d, you get a parallel plane. So if you generate d for one point, you can check it with another (both of the others, if you're nervous ).

## Finding the equation of a plane passing through 3 points

Oh man, can't believe I didn't see that. Thanks!