Question

Three consecutive 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 D.

Answer 1

Given that, a parallelogram $ABCD$ has three consecutive vertices as $A(3,-1,2),B(1,2,-4)$ and $C(-1,1,2)$.

Let the coordinates of the fourth vertex $D$ be $(x,y,z)$

We know that, the diagonals of a parallelogram bisect each other.

So, if $X$ be the point of intersection of both diagonals, $AC$ and $BD$, it will be the midpoint of both $AC$ and $BD$.

We also know that the coordinates of the midpoint of a line joining two points with coordinates $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2})$

$\therefore$Midpoint of $AC=(\frac{3-1}{2},\frac{-1+1}{2},\frac{2+2}{2})=(1,0,2)$

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

Since, midpoint of $AC$ and $BD$ is $X$,

$\therefore (\frac{1+x}{2},\frac{2+y}{2},\frac{-4+z}{2})=(1,0,2)$

$\Rightarrow \frac{1+x}{2}=1,\frac{2+y}{2}=0,\frac{-4+z}{2}=2$

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

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

Hence, the coordinates of the fourth vertex $D$ are $(1,-2,8)$