Area of triangle created from 3 planes

AI Thread Summary
The discussion centers on finding the area of a triangle formed by the intersection of three non-parallel planes in a 3D space. The planes are defined by vector equations, and the triangle is created by a plane perpendicular to the line formed by the intersection of two of the planes. Participants suggest starting with the area formula for a triangle using vectors and emphasize the need to express the area in terms of specific scalar quantities related to the planes. The solution involves determining the coordinates of the triangle's vertices and calculating the area using vector cross products. Ultimately, the challenge lies in converting the area expression into the required scalar terms.
TOD
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Homework Statement


Ok, given 3 planes pi1, pi2 and pi3 with vector equations r.n1=0, r.n2=0 and (r-a).n3=0 respectively, where a, n1, n2, n3 are given vectors. No 2 planes are parallel and the third plane is parallel to the line, L, given by the intersection of planes pi1 and pi2. Consider the triangle obtained by intersection of all three planes by a plane perpendicular to L. Find a formula for the area and express it only in terms of the scalars: a.n3, |n1xn2|, |n1xn3| and |n2xn3|.


Homework Equations


See above


The Attempt at a Solution


Too many failed attempts... But I guess start of with or end up with some sort of equation that looks like A=0.5|uxv| (i.e. area of a triangle using vectors).
I also believe planes pi1 and pi2 have a common point at (0,0,0). I don't know how to find other common points without turning things into lots of variables via cartesian equations.

I've beens stuck on this question for about 12 hours (not consecutively of coures) and still can't figure out the answer... Any help will be appreciated!
Thanks in advance,
- TOD
 
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Do you mind "cheating"?

We can, without loss of generality, set up our coordinate system so that pi1 is the yz-plane, x= 0: that is n1= (1, 0, 0). A general plane, having the z-axis as intersection with that, is Ax+ By= 0 and we can take pi2 to be that: n2= (A, B, 0). Finally, since pi3 must be parallel to L, the z axis, pi3 can be taken to be Px+ Qy+ R= 0: n3= (P, Q, 0) and a= (-R/P, 0, 0). Finally, we can take the fourth plane, "perpendicular to L", to be the xy-plane.

It's easy to see that one vertex of the triangle, where pi1, pi2, and the "fourth plane" intersect, is just (0, 0, 0).
It's also easy to solve for the intersection of pi2 and pi3 (remembering that z= 0 in all this) and the intersection of pi1 and pi3 to get the coordinates of the other two vertices of the triangle. Once you have those, you can write u and v as the vectors from (0,0,0) to those points, and calculate (1/2)||uxv|| to find the area of the triangle, in terms of A, B, P, Q, and R.

Now, the hard part! You need to convert that answer to "a.n3, |n1xn2|, |n1xn3| and |n2xn3|". You should have got a fraction as the area- and the denominator is obvious. You need to find how to write the numerator as one of those vector products. (Well, that's not as hard as I thought it was!)
 
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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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