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

[mathmari]

Hello!

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)

 

[Dethrone]

Yes, that is correct, using the fact that the area of a parallelogram can be calculated by the magnitude of its cross product. Regarding justifying that the area of a triangle is half the area of a parallelogram I think requires a geometric proof: https://proofwiki.org/wiki/Parallelogram_on_Same_Base_as_Triangle_has_Twice_its_Area

 

[pickslides]

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)

Sounds like a good idea. Follow these steps.

3-Coordinate Triangle Calculator | Area/Perimeter/Angle

Problem on the area of a triangle in 3-space - Leading Lesson

 

[Prove It]

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 :)

 

[MarkFL]

You could also use the formula developed here:

http://mathhelpboards.com/math-notes-49/finding-area-triangle-formed-3-points-plane-2954.html