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Question
Show that the points A(3, 3, 3), B(0, 6, 3), C(1, 7, 7) and D(4, 4, 7) are the vertices of a square.
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Solution
Let A(3,3,3) , B(0,6,3) , C( 1,7,7) and D (4,4,7) are the vertices of quadrilateral \[\square ABCD\]
We have :
AB =\[\sqrt{\left( 0 - 3 \right)^2 + \left( 6 - 3 \right)^2 + \left( 3 - 3 \right)^2}\]
\[ = \sqrt{9 + 9 + 0}\]
\[ = \sqrt{18}\]
\[ = 3\sqrt{2}\]
BC =\[\sqrt{\left( 1 - 0 \right)^2 + \left( 7 - 6 \right)^2 + \left( 7 - 3 \right)^2}\]
\[= \sqrt{\left( 1 \right)^2 + \left( 1 \right)^2 + \left( 4 \right)^2}\]
\[ = \sqrt{1 + 1 + 16}\]
\[ = \sqrt{18}\]
\[ = 3\sqrt{2}\]
CD =\[\sqrt{\left( 4 - 1 \right)^2 + \left( 4 - 7 \right)^2 + \left( 7 - 7 \right)^2}\]
\[= \sqrt{\left( 3 \right)^2 + \left( - 3 \right)^2 + \left( 0 \right)^2}\]
\[ = \sqrt{9 + 9 + 0}\]
\[ = \sqrt{18}\]
\[ = 3\sqrt{2}\]
DA =\[\sqrt{\left( 4 - 3 \right)^2 + \left( 4 - 3 \right)^2 + \left( 7 - 3 \right)^2}\]
\[ = \sqrt{1 + 1 + 16}\]
\[ = \sqrt{18}\]
\[ = 3\sqrt{2}\]
AB = BC = CD = DA
AC =
AB = BC = CD = DA
AC =\[\sqrt{\left( 1 - 3 \right)^2 + \left( 7 - 3 \right)^2 + \left( 7 - 3 \right)^2}\]
\[ = \sqrt{4 + 16 + 16}\]
\[ = \sqrt{36}\]
\[ = 6\]
\[\]
\[ = \sqrt{16 + 4 + 16}\]
\[ = \sqrt{36}\]
\[ = 6\]
Therefore, the points are the vertices of a square.
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