Question

Here are the vertices of my regular tetrahedron, taken from a cube centered at (0,0,0)
\[A:=(1,1,1)\]
\[B:=(−1,1,−1)\]
\[C:=(−1,−1,1)\]
\[D:=(1,−1,−1)\]

what is the area of one face of that tetrahedron?

Answer 1

The area corresponds to the determinant, or to the cross product, between 2 adjacent edge vectors.

Answer 2

I will consider 2 adjacent vectors\[AB = (-2,0,-2)\]
\[AC = (-2,-2,0)\]

Answer 3

what is their cross product?

Answer 4

it is det(
i j k
-2 0 -2
-2 -2 0
)

which is -4(i-j-k)

Answer 5

or (-4, 4, 4)

what is the length of that vector

Answer 6

\[4\sqrt{3}\]
what is half of that?

\[2\sqrt{3}\]

that is the area of one face of my regular tetrahedron taken from a cube with side length 2

Answer 7

is that right? I think I am wrong

Answer 8

just find the magnitude of say AB then use trig or pythagorean theorem to find the height of a face then do 1/2 base times height

Answer 9

the length of (-4,4,4) would be sqrt(4^2+4^2+4^2)

Answer 10

hmm...

\[12 = 24 - 4\sqrt{3}*\frac{1}{2}\]doesn't sound right

Answer 11

but that's because I wrote it wrong.

\[12=24-2(4*3)*0.5\]

Answer 12

yeah, \[2\sqrt{3}\]is the area of the side! thanks

Answer 13

It's a regular tetrahedron, so the faces are all equilateral triangles. The sides of the triangles are diagonals of the cube faces, so sqrt(8). In an 30-60-90 right triangle (half of an equilateral triangle) the sides are in the ratio 1-sqrt(3)-2. So the height of one of the triangular faces will be sqrt(3)*sqrt(8)/2 = sqrt(6). Then the area will be 1/2sqrt(6)sqrt(8) = 1/2sqrt(48) = 2sqrt(3).

The total area of the tetrahedron must be less than the total are of the cube.
4(2sqrt(3)) = 8*1.732 = 13.856
6(4) = 24
So that checks out.