Show that $A B C D$ is a square where $A, B, C, D$ are the points $(0, 4, 1), (2, 3, - 1), (4, 5, 0)$ and $(2, 6, 2)$ respectively.

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Solution

Given $A (0, 4, 1) B (2, 3, - 1) C (4, 5, 0) D (2, 6, 2)$
$A B = \sqrt{(0 - 2)^{2} + (4 - 3)^{2} + (1 + 1)^{2}} = 3$
$B C = \sqrt{(2 - 4)^{2} + (3 - 5)^{2} + (- 1 - 0)^{2}} = 3$
$C D = \sqrt{(4 - 2)^{2} + (5 - 6)^{2} + (0 - 2)^{2}} = 3$
$D A = \sqrt{(2 - 0)^{2} + (6 - 4)^{2} + (2 - 1)^{2}} = 3$
$∴ A B = B C = C D = D A$
$A C = \sqrt{(0 - 4)^{2} + (4 - 5)^{2} + (1 - 0)^{2}} = \sqrt{18}$
$B D = \sqrt{(2 - 2)^{2} + (3 - 6)^{2} + (- 1 - 2)^{2}} = \sqrt{18}$
$∴ A C = B C$ .
$∴ A B^{2} + B C^{2} = A C^{2}$
$9 + 9 = 18$ .
$∠ A B C = 90^{o}$
$∴ A, B, C, D$ are vertices of square.

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