1

Say I wanted to calculate the area of a figure comprised of the coordinates (0,1,1) (1,0,1) and (0,0,0). When I used cross products I get the area as the square root of 3/2, however, when I use standard geometry I get the answer as 1/6. Can someone please point out my mistake when using cross products( I got the vectors as <0,1,1> and <-1,1,0> and I got the cross product as <-1,1,1>

user501887
  • 35
  • What is this "standard geometry"? – Angina Seng Dec 31 '18 at 19:06
  • Since the pyramid has base area 1/2 and height 1 the area of the pyramid would be 0.5* 1/3*1 =1/6 – user501887 Dec 31 '18 at 19:11
  • 1
    Pyramid? You have three vertices; those define a triangle for me, not a pyramid. – Angina Seng Dec 31 '18 at 19:13
  • I might be wrong, but I am pretty sure that we can't form a triangle in the 3d plane – user501887 Dec 31 '18 at 19:16
  • I think I can, and this one has area $\sqrt3/2$. – Angina Seng Dec 31 '18 at 19:20
  • 1
    @user501887 to have a pyramid in 3d, you have to have 4 vertices. You only have three. There is no shape with non-zero area in the euclidean geometry that has three vertices, other than a tringle, no matter what dimension you talk about – KKZiomek Dec 31 '18 at 20:45
  • 1
    If the figure were described as in your above comment —"[a] pyramid [having] base area 1/2 and height 1"— then $0.5\cdot\frac13\cdot 1=\frac16$ wouldn't be its area, but its volume. That said, as others have commented, three points make a triangle, not a pyramid. To your comment about our (in)ability to form triangles "in the 3d plane": every triangle you have ever encountered in your life exists in 3d. :) – Blue Dec 31 '18 at 22:35
  • What makes you think that it’s your computation via a cross product that produced the wrong result? – amd Jan 01 '19 at 00:31

2 Answers2

1

Here is what my standard geometry tells me. You have an isosceles triangle with two sides of length $\sqrt2$ and the angle between them is $\cos^{-1}(1/2) =\pi/3$. Its area is $$\frac12(\sqrt2)^2\sin\frac\pi3=\frac{\sqrt3}2.$$

Angina Seng
  • 158,341
  • You mean you have an equilateral triangle :) And I would use Pythagoras' theorem, otherwise the formula looks too much like the cross product. – Andrei Dec 31 '18 at 19:51
1

The sides are given by vectors $(1,0,1), (0,1,1), (1,-1,0).$ The triangle is equilateral with lengths of sides $a=\sqrt 2$ and the area $$\mathcal{A}=\frac{a}{2}\cdot\frac{a\sqrt3}{2}=\frac{\sqrt3}{2}$$

user376343
  • 8,311