Question

Three consecutive vertices of a parallelogram $ABCD$ are $A(-2,3,5),B(1,2,4)$ and $C(3,5,1)$.Find the fourth vertex $D$

Answer 1

In a parallelogram $ABCD$ are $A(-2,3,5),B(-3,2,1)$ and $C(3,5,1)$.Let the fourth vertex $D$ be $(x,y,z)$

Midpoint of $AC=(\frac{-2+3}{2},\frac{3+5}{2},\frac{5+1}{2})=(\frac{1}{2},\frac{8}{2},\frac{6}{2})$

Midpoint of $BD=(\frac{1+x}{2},\frac{2+y}{2},\frac{4+z}{2})$

Since $ABCD$ is a parallelogram and in a parallelogram, the diagonals bisect each other, so the mid-points of $AC$ and $BD$ are same,

$\therefore (\frac{1+x}{2},\frac{2+y}{2},\frac{4+z}{2})=(\frac{1}{2},\frac{8}{2},\frac{6}{2})$

$\Rightarrow \frac{1+x}{2}=\frac{1}{2},\frac{2+y}{2}=\frac{8}{2},\frac{4+z}{2}=\frac{6}{2}$

$\Rightarrow x+1=1,y+2=8,z+4=6$

$\Rightarrow x=1-1=0,y=8-2=6,z=6-4=2$