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Area of triangle created from 3 planes

  1. Apr 14, 2008 #1

    TOD

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    1. The problem statement, all variables and given/known data
    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|.


    2. Relevant equations
    See above


    3. 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 dunno 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
     
  2. jcsd
  3. Apr 14, 2008 #2

    HallsofIvy

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    Staff Emeritus
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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!)
     
  4. Apr 14, 2008 #3

    TOD

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    Wow Genius! Thank Youuuuuuuuuu! Love You Mannnn! Love You Soooo Sooo Much!!! ^_______^
     
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