Area of triangle created from 3 planes

Click For Summary
SUMMARY

The discussion focuses on deriving a formula for the area of a triangle formed by the intersection of three planes defined by vector equations. The planes are represented as pi1, pi2, and pi3, with specific normal vectors n1, n2, and n3. The area is calculated using the formula A = 0.5 |u x v|, where u and v are vectors from the origin to the triangle's vertices. The final expression for the area is to be expressed in terms of the scalars a.n3, |n1 x n2|, |n1 x n3|, and |n2 x n3|.

PREREQUISITES
  • Vector calculus and operations, specifically cross products
  • Understanding of plane equations in three-dimensional space
  • Knowledge of linear algebra concepts, particularly vector representation
  • Familiarity with geometric interpretations of vector operations
NEXT STEPS
  • Study vector cross product properties and applications in geometry
  • Learn about the geometric interpretation of planes and their intersections
  • Explore methods for calculating areas of triangles in three-dimensional space
  • Investigate the relationship between vector equations and scalar products
USEFUL FOR

Students studying advanced geometry, mathematics enthusiasts, and anyone tackling problems involving vector equations and plane intersections.

TOD
Messages
18
Reaction score
0

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
 
Physics news on Phys.org
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!)
 
Wow Genius! Thank Youuuuuuuuuu! Love You Mannnn! Love You Soooo Sooo Much! ^_______^
 

Similar threads

Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K