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.

Answer 1

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 =\[\sqrt{\left( 0 - 3 \right)^2 + \left( 6 - 3 \right)^2 + \left( 3 - 3 \right)^2}\] 

\[= \sqrt{\left( - 3 \right)^2 + \left( 3 \right)^2 + \left( 0 \right)^2}\]
\[ = \sqrt{9 + 9 + 0}\]
\[ = \sqrt{18}\]
\[ = 3\sqrt{2}\]

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 =\[\sqrt{\left( 4 - 3 \right)^2 + \left( 4 - 3 \right)^2 + \left( 7 - 3 \right)^2}\]

\[= \sqrt{\left( 1 \right)^2 + \left( 1 \right)^2 + \left( 4 \right)^2}\]
\[ = \sqrt{1 + 1 + 16}\]
\[ = \sqrt{18}\]
\[ = 3\sqrt{2}\]

AB = BC = CD = DA

AC =
AB = BC = CD = DA

AC =\[\sqrt{\left( 1 - 3 \right)^2 + \left( 7 - 3 \right)^2 + \left( 7 - 3 \right)^2}\]

\[= \sqrt{\left( - 2 \right)^2 + \left( 4 \right)^2 + \left( 4 \right)^2}\]
\[ = \sqrt{4 + 16 + 16}\]
\[ = \sqrt{36}\]
\[ = 6\] 

BD =\[\sqrt{\left( 4 - 0 \right)^2 + \left( 4 - 6 \right)^2 + \left( 7 - 3 \right)^2}\]
\[\]

\[= \sqrt{\left( 4 \right)^2 + \left( - 2 \right)^2 + \left( 4 \right)^2}\]
\[ = \sqrt{16 + 4 + 16}\]
\[ = \sqrt{36}\]
\[ = 6\]

\[\therefore\]AC = BD 

Since, all sides and diagonals of quadrilateral\[\square ABCD\] are equal
Therefore, the points are the vertices of a square.