Given: Points are A(1,2, 3), B(-1, -2, -1), C(2, 3, 2) and D(4, 7, 6)
To prove: the quadrilateral formed by these 4 points is a parallelogram but not a rectangle
Opposite sides of both parallelogram and rectangle are equal
But diagonals of a parallelogram are not equal whereas they are equal for rectangle
Formula used:
The distance between any two points (a, b, c) and (m, n, o) is given by,
\(\sqrt{(a-m)^2+(b-n)^2+(c-o)^2}\)
Therefore,
Distance between A(1, 2, 3) and B(-1, -2, -1) is AB,
Distance between B(-1, -2, -1) and C(2, 3, 2) is BC,
Distance between C(2, 3, 2) and D(4, 7, 6) is CD,
The distance between A(1, 2, 3) and D(4, 7, 6) is AD,
Clearly, AB = CD
BC = AD
Opposite sides are equal
Now, we will find the length of diagonals
The distance between A(1, 2, 3) and C(2, 3, 2) is AC,
=\(\sqrt{(1-2)^2+(2-3)^2+(3-2)^2}\)
= \(\sqrt{(-1)^2+(-1)^2+1^2}\)
= \(\sqrt{1+1+1}\)
= \(\sqrt{3}\)
Distance between B(-1, -2, -1) and D(4, 7, 6) is BD,
=\(\sqrt{(-1-4)^2+(-2-7)^2+(-1-6)^2}\)
= \(\sqrt{(-5)^2+(-9)^2+(-5)^2}\)
= \(\sqrt{25+81+25}\)
= \(\sqrt{131}\)
Clearly,
AC ≠ BD
The diagonals are not equal, but opposite sides are equal
Thus, Quadrilateral formed by ABCD is a parallelogram but not a rectangle
Hence Proved
