Question

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

Answer 1

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

$(x,y,z)$

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

Midpoint of $BD=(\frac{-3+x}{2},\frac{2+y}{2},\frac{1+z}{2})$

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

$\therefore (\frac{-3+x}{2},\frac{2+y}{2},\frac{1+z}{2})=(\frac{-3}{2},\frac{-1}{2},4)$

$\Rightarrow \frac{-3+x}{2}=\frac{-3}{2},\frac{2+y}{2}=\frac{-1}{2},\frac{1+z}{2}=4$

$\Rightarrow x-3=-3,y+2=-1,z+1=8$

$\Rightarrow x=-3+3=0,y=-1-2=-3,z=8-1=7$

Hence, the fourth vertex $D$ be $(0,-3,7)$