Question

Show that the points $A(1,2,3),B(-1,-2,-3),C(2,3,2)$ and $D(4,7,6)$ are the vertices of a parallelogram.

Answer 1

To prove ; $ABCD$ is a parallelogram for this we'll show that both the diagonal meet at $0$

i.e., $0$ is the midpoint of $AC$ and $BD$

$\left(BD\right)0\equiv (\frac{-1+4}{2},\frac{-2+7}{2},\frac{-1+6}{2})$

$\left(AC\right)0\equiv (\frac{1+2}{2},\frac{2+3}{2},\frac{3+2}{2})$

$0\equiv (\frac{3}{2},\frac{5}{2},\frac{5}{2})0\equiv (\frac{3}{2},\frac{5}{2},\frac{5}{2})$

Hence, $A(1,2,3),B(-1,-2,-1),C(2,3,2)$ and $D(4,7,6)$ are the vertices of a parallelogram.