Tetrahedron Vertex Coordinates
The points $ A = (-1,1,1), $ $ B = (1,-1,1), $ $ C = (1,1,-1), $ and $ D = (-1,-1,-1) $ form the vertices of a regular tetrahedron. Find the coordinates of the point $ E, $ other than $ D, $ such that $ ABCE $ is also a regular tetrahedron.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
The side length of regular tetrahedron $ ABCE $ is equal to the distance between $ A $ and $ B, $ which is $ \sqrt{2^2 + 2^2} = 2 \sqrt{2} $.
Let $ E = (x,y,z) $. Since $ ABCE $ is a regular tetrahedron, we want $ AE = BE = CE = 2 \sqrt{2} $. Thus,
\begin{align*}
(x + 1)^2 + (y - 1)^2 + (z - 1)^2 &= 8, \\
(x - 1)^2 + (y + 1)^2 + (z - 1)^2 &= 8, \\
(x - 1)^2 + (y - 1)^2 + (z + 1)^2 &= 8.\end{align*}Subtracting the first two equations, we end up with $ x = y $. Subtracting the first and third equations, we end up with $ x = z $. Then we can write the first equation as
\[(x + 1)^2 + (x - 1)^2 + (x - 1)^2 = 8.\]This simplifies to $ 3x^2 - 2x - 5 = 0, $ which factors as $ (x + 1)(3x - 5) = 0 $. Hence, $ x = -1 $ or $ x = \frac{5}{3} $.
If $ x = -1, $ then $ E $ will coincide with $ D, $ so $ x = \frac{5}{3}, $ which means $ E = \boxed{\left( \frac{5}{3}, \frac{5}{3}, \frac{5}{3} \right)} $.
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