Need to find the coordinates of the top points of a cube, A, B, C, and D

**Ants** · August 21st 2012, 01:38 AM

Given information: A(13,30,9) C(2,5,17) B(bx,by,13) D(dx,dy,13), hence the height of points B and D are the same .
Need to determine bx,by,dx,dy.
Segment AC and BD are the diagonals.

Thanks

**Soroban** · August 21st 2012, 12:56 PM

Hello, Ants!

I have a game plan . . .

$A(13,30,9),\;B(x_b,y_b,13),\lC(2,5,17),\;D(x_d,y_d ,13)\,\text{ form a square.}$
$\text{Segments }AC\text{ and }BD\text{ are the diagonals.}$
$\text{Determine }x_b,y_b,x_d,y_d.$

Code:

                  (2,5,17)
                      ♥ C
                  *    *
          D   *         *
(x_d,y_d,13) ♥              *
           *              *
            *              *
             *              ♥ (x_b,y_b,13)
              *         *   B
               *    *
              A ♥
            (13,30,9)

We have four unknowns; we will need four equations.

The sides of the square are equal.

. . $\begin{array}{ccccc} \overline{AB}^2 &=& (x_b-13)^2 + (y_b-30)^2 + (13-9)^2 \\ \overline{BC}^2 &=& (x_b-2)^2 + (y_b-5)^2 + (13-17)^2 \\ \overline{CD}^2 &=& (x_d-2)^2 + (y_d-5)^2+(13-17)^2 \\ \overline{DA}^2 &=& (x_d-13)^2 + (y_d-30)^2 + (13-9)^2 \end{array}$

These give us three equations.

The diagonals are equal.

. . $\overline{AC}^2 \:=\:11^2 + 25^2 + 8^2 \:=\:810 \quad\Rightarrow\quad AC \:=\:9\sqrt{10}$

. . $\overline{BD}^2 \:=\:(x_d-x_b)^2 + (y_d-y_b)^2 + (13-13)^2$

This gives us the fourth equation.

Can you take it from here?

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Need to find the coordinates of the top points of a cube, A, B, C, and D

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