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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.

 Apr 29, 2020
 #1
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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)

 

cool cool cool

 Apr 30, 2020

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