Question

Three consecutive vertices of a parallelogram $ABCD$ are $A(1,-2,3),B(1,2,4)$ and $C(3,3,6)$.Find the fourth vertex $D$

Answer 1

In a parallelogram $ABCD$ are $A(1,-2,3),B(1,2,4)$ and $C(3,3,6)$.Let the fourth vertex $D$ be $(x,y,z)$

Midpoint of $AC=(\frac{1+3}{2},\frac{-2+3}{2},\frac{3+6}{2})=(\frac{4}{2},\frac{1}{2},\frac{9}{2})$

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,

$\therefore (\frac{1+x}{2},\frac{2+y}{2},\frac{4+z}{2})=(\frac{4}{2},\frac{1}{2},\frac{9}{2})$

$\Rightarrow \frac{1+x}{2}=\frac{4}{2},\frac{2+y}{2}=\frac{1}{2},\frac{4+z}{2}=\frac{9}{2}$

$\Rightarrow x+1=4,y+2=1,z+4=9$

$\Rightarrow x=4-1=3,y=1-2=-1,z=9-4=5$

Hence, the fourth vertex $D$ be $(3,-1,5)$