Finding the equation of a plane passing through 3 points

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To find the equation of a plane through points A(2,1,0), B(3,-1,5), and C(2,2,1), the normal vector is calculated using the cross product of vectors AB and AC, resulting in n=(-7, -1, 1). The equation of the plane can be expressed as -7x - y + z = d, but the correct form is 7x + y - z = 15, which involves sign inversion. To determine the value of d, one can substitute the coordinates of any of the three points into the plane equation. This process confirms that the derived equation represents the same plane, ensuring consistency across all points. Understanding the sign inversion and solving for d is crucial for accurately representing the plane's equation.
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Homework Statement


Find the equation of the plane passing through point A(2,1,0), B(3,-1,5) and C(2,2,1)


Homework Equations


Um..I don't know?


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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Luscinia said:

Homework Statement


Find the equation of the plane passing through point A(2,1,0), B(3,-1,5) and C(2,2,1)


Homework Equations


Um..I don't know?


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 :smile:).
 
Oh man, can't believe I didn't see that. Thanks!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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