MHB Find Area of Triangle with Vertices $(0, 0, 0), (1, 1, 1)$ and $(0, -2, 3)$

  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    Area Triangle
mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hello! :o

We have a triangle with vertices $(0, 0, 0), (1, 1, 1)$ and $(0, -2, 3)$. We want to find the area.

How could we find it?? Do we maybe use the fact that the area of the triangle is the half of the area of the parallelogram?? (Wondering)

How do we know that it stands?? How can we justify it?? (Wondering)
 
Physics news on Phys.org
Another option could be to work out the length of each segment (using Pythagoras), then the area can be found using Heron's Formula.

But the vector method is much quicker :)
 
You could also use the formula developed here:

http://mathhelpboards.com/math-notes-49/finding-area-triangle-formed-3-points-plane-2954.html
 
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1 Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers. So we only handle with original Zeta...
Back
Top