Require to show that the given co-planar points A(2,3,2), B(4,7,6), C(1,2,3), and D(-1,-2,-1)form a parallelogram.
Find the lengths of the sides AB, BC, CD and AD of the quadrilateral A B C D ABCD A BC D
Now
A B = ( 4 − 2 ) 2 + ( 7 − 3 ) 2 + ( 6 − 2 ) 2 = 4 + 16 + 16 = 6 AB=\sqrt{(4-2)^2+(7-3)^2+(6-2)^2}=\sqrt{4+16+16}=6 A B = ( 4 − 2 ) 2 + ( 7 − 3 ) 2 + ( 6 − 2 ) 2 = 4 + 16 + 16 = 6
B C = ( 1 − 4 ) 2 + ( 2 − 7 ) 2 + ( 3 − 6 ) 2 = 9 + 25 + 9 = 43 BC=\sqrt{(1-4)^2+(2-7)^2+(3-6)^2}=\sqrt{9+25+9}=\sqrt{43} BC = ( 1 − 4 ) 2 + ( 2 − 7 ) 2 + ( 3 − 6 ) 2 = 9 + 25 + 9 = 43
C D = ( − 1 − 1 ) 2 + ( − 2 − 2 ) 2 + ( − 1 − 3 ) 2 = 4 + 16 + 16 = 6 CD=\sqrt{(-1-1)^2+(-2-2)^2+(-1-3)^2}=\sqrt{4+16+16}=6 C D = ( − 1 − 1 ) 2 + ( − 2 − 2 ) 2 + ( − 1 − 3 ) 2 = 4 + 16 + 16 = 6
A D = ( − 1 − 2 ) 2 + ( − 2 − 3 ) 2 + ( − 1 − 2 ) 2 = 9 + 25 + 9 = 43 AD=\sqrt{(-1-2)^2+(-2-3)^2+(-1-2)^2}=\sqrt{9+25+9}=\sqrt{43} A D = ( − 1 − 2 ) 2 + ( − 2 − 3 ) 2 + ( − 1 − 2 ) 2 = 9 + 25 + 9 = 43
Observe that A B = C D AB=CD A B = C D and B C = A D BC=AD BC = A D
That is opposite sides of the quadrilateral A B C D ABCD A BC D are equal.
That is the quadrilateral A B C D ABCD A BC D is a parallelogram.
Therefore, the given co-planar points form a parallelogram.
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