Three vertices of a cube in space are A = (2,3,0), B = (0,5,4), and C = (4,1,8). Find the center of the cube.
+128474 Call the center (a, b, c)
The distance from any vertex to the center is equal.....so.......we can generate this system of equations :
(a- 2)^2 + ( b - 3)^2 + (c - 0)^2 = (a-0)^2 + (b - 5)^2 + (c - 4)^2
a^2 - 4a + 4 +b^2 - 6b + 9 + c^2 = a^2 + b^2 -10b + 25 + c^2 - 8c + 16
-4a + 4 - 6b + 9 = -10b + 25 - 8c + 16
-4a + 4b + 8c = 28
-a + b + 2c = 7 (1)
(a -2)^2 + (b-3)^2 + (c-0)^2 = (a-4)^2 + (b -1)^2 + (c -8)^2
a^2 - 4a + 4 + b^2 - 6b + 9 + c^2 = a^2 - 8a + 16 + b^2 -2b + 1 + c^2 - 16c + 64
4a - 4b + 16c = 68
a - b + 4c = 17 (2)
(a-0)^2 + (b - 5)^2 + (c - 4)^2 = (a-4)^2 + (b -1)^2 + (c -8)^2
a^2 + b^2 - 10b + 25 + c^2 - 8c + 16 = a^2 - 8a + 16 + b^2 -2b + 1 + c^2 - 16c + 64
8a -8a + 8c = 40
a - b + c = 5 (3)
Add (1) and ( 2) and we get that
6c = 24
c = 4
Subbing this into (1) for c we have that
-a + b + 8 = 7 ⇒ -a + b = -1 ⇒ a - b = 1 ⇒ b = a - 1
The center is ( a , a - 1, 4)
