Question

Prove that the point A(1, 3, 0), B(-5, 5, 2), C(-9, -1, 2) and D(-3, -3, 0) taken in order are the vertices of a parallelogram. Also, show that ABCD is not a rectangle.

Answer 1

Here, AB=$\sqrt{(5+1{)}^2+(5-3{)}^2+(2-0{)}^2}$

=$\sqrt{36+4+4}$

=$\sqrt{44}$

=$2\sqrt{11}$

BC=$\sqrt{(-9+5{)}^2+(-1-5{)}^2+(2-2{)}^2}$

=$\sqrt{16+36}$

=$\sqrt{52}$

=2$\sqrt{13}$

CD=$\sqrt{(-3+9{)}^2+(-3+1{)}^2+(0-2{)}^2}$

=$\sqrt{36+4+4}$

=2$\sqrt{11}$

DA=$\sqrt{(1+3{)}^2+(3+3{)}^2+0}$

=$\sqrt{16+36}$

=$\sqrt{52}$

=2$\sqrt{13}$

AC=$\sqrt{(-9-1{)}^2+(-1-3{)}^2+(2-0{)}^2}$

=$\sqrt{150+16+4}$

=$\sqrt{120}$

=4$\sqrt{5}$

BD=$\sqrt{(-3+5{)}^2+(-3-5{)}^2+(0-2{)}^2}$

=$\sqrt{4+64+4}$

=$\sqrt{72}$

=6$\sqrt{2}$

Since,

AB = CD and BC = DA

$\Rightarrow$ ABCD is a parallelogram

but, AC $\ne$ BD

$\Rightarrow$ ABCD is not a rectangle.