Question

Find the area of the triangle with vertices (1,1,5) (3,4,3) and (1,5,7)

Answer 1

I got 8.307 am I right?

Answer 2

what is the perimeter?

Answer 3

For (Xa,Xb,Xc)=(1,1,5),(Ya,Yb,Yc)= (3,4,3)and (Za,Zb,Zc)=(1,5,7):
http://upload.wikimedia.org/math/1/7/c/17cb8db9ccd42b47a704ba683c7c5f5b.png
Where | | is the determinant.

Answer 4

Correction:
For (Xa,Ya,Za)=(1,1,5),(Xb,Yb,Zb)= (3,4,3)and (Xc,Yc,Zc)=(1,5,7):
\(\frac{1}{2}\sqrt{(-4)^{2}+8^{2}+14^{2}}=\sqrt{69} \approx \) 8.307

Answer 5

Area of triangle = 1/2 * |axb|
Let a be the vector from (1,1,5) to (3,4,3) Let b be the vector from (3,4,3) to (1,5,7)

vector a = (1,1,5) - (3,4,3) = (-2,-3,2)
vector b = (3,4,3) - (1,5,7)= (2,-1,-4)

After that, find the cross product a x b

| i j k |
| -2i,-3j,2k|
| 2i,-1j,-4k|

=(14i,-4j,8k)

since you’ve got the answer for the cross product now find the area of parallelogram of vector b and a.
|a x b| = sqrt( (14)^2 + (-4)^2 + (8)^2)= sqrt 69
area of triangle = 1/2 * sqrt (69)=8.307!

Thanks @ybarrap , i did something wrong with my calculations :)

Answer 6

Your solution is a lot more elegant. Amazing the number of ways to find areas!