Question

Show that the points A(1, 2, 3), B(–1, –2, –1), C(2, 3, 2) and D(4, 7, 6) are the vertices of a parallelogram ABCD, but not a rectangle.

Answer 1

To show that ABCD is a parallelogram, we need to show that its two opposite sides are equal.

\[AB = \sqrt{\left( - 1 - 1 \right)^2 + \left( - 2 - 2 \right)^2 + \left( - 1 - 3 \right)^2}\]
\[ = \sqrt{4 + 16 + 16}\]
\[ = \sqrt{36}\]
\[ = 6\]
\[BC = \sqrt{\left( 2 + 1 \right)^2 + \left( 3 + 2 \right)^2 + \left( 2 + 1 \right)^2}\]
\[ = \sqrt{9 + 25 + 9}\]
\[ = \sqrt{43}\]
\[CD = \sqrt{\left( 4 - 2 \right)^2 + \left( 7 - 3 \right)^2 + \left( 6 - 2 \right)^2}\]
\[ = \sqrt{4 + 16 + 16}\]
\[ = \sqrt{36}\]
\[ = 6\]
\[DA = \sqrt{\left( 1 - 4 \right)^2 + \left( 2 - 7 \right)^2 + \left( 3 - 6 \right)^2}\]
\[ = \sqrt{9 + 25 + 9}\]
\[ = \sqrt{43}\]
\[AB = CD and BC = DA\]
\[\text{ Since, opposite pairs of sides are equal } . \]
\[ \therefore \text{ ABCD is a parallelogram }\]
\[AC = \sqrt{\left( 2 - 1 \right)^2 + \left( 3 - 2 \right)^2 + \left( 2 - 3 \right)^2}\]
\[ = \sqrt{1 + 1 + 1}\]
\[ = \sqrt{3}\]
\[BD = \sqrt{\left( 4 + 1 \right)^2 + \left( 7 + 2 \right)^2 + \left( 6 + 1 \right)^2}\]
\[ = \sqrt{25 + 81 + 49}\]
\[ = \sqrt{155}\]
\[\text{ Since }, AC \neq BD\]
Thus, ABCD is not a rectangle.