Question

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

Answer 1

Let $D(x,y,z)$ be the fourth vertex of parallelogram ABCD.

We know that diagonals of a parallelogram bisect each other. So the mid points of AC and BD coincide.

$\therefore$ Coordinates of mid point of AC

$(\frac{3-1}{2},\frac{-1+1}{2},\frac{2+2}{2})$

= $(1,0,2)$

Also coordiantes of mid point of BD

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

$\therefore \frac{x+1}{2}=1\Rightarrow x+1=2\Rightarrow x=1$

$\frac{y+2}{2}=0\Rightarrow y+2=0\Rightarrow y=-2$

$\frac{z-4}{2}=2\Rightarrow z-4=4\Rightarrow z=8$

Thus the coordinates of point D are $(1,-2,8).$