Finding the Equation of a Plane from 3 Points

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I have a general question. If given 3 points, how would I find the equation of the plane containing all these points?
 
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Do you have any ideas?

Do you know of any ways to write the equation of a plane given some other information? Can you gather that information if given 3 points?
 
I think I found it. I would need to take the cross product of 2 vectors from those 3 points to find the normal vector. Then I would use the equation 0 = a(x-x1)+b(y-y1)+c(z-z1), <a,b,c> being the normal vector. For <x1,y1,z1> would I just pick one of the points? Does it matter which one?
 
Your approach is fine and it doesn't matter which point you use in the end.
 
That form of the plane equation works but I prefer to write the plane equation as ax+by+cz=d where <a,b,c> is a normal vector to the plane. You can then solve for d by evaluating the left-hand side at any point. In the end it doesn't matter because you'll end up with the same equation.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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