Question

Let P(1,2,1), Q(1,0,-1), R(2,2,0) be the vertices of a parallelogram with adjacent sides PQ and PR. Find the other vertex S.

Answer 1

Given:

The vertices of a parallelogram are P(1,2,1), Q(1,0,-1), R(2,2,0).

PQ and PR are the adjacent sides of the parallelogram.

To find:

The coordinates of vertex S.

Solution:

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

Let the coordinates of the vertex S are (a,b,c).

In the given parallelogram PS and QR are the diagonals. It means their midpoints are same.

$\left(\dfrac{1+a}{2},\dfrac{2+b}{2},\dfrac{1+c}{2}\right)=\left(\dfrac{1+2}{2},\dfrac{0+2}{2},\dfrac{-1+0}{2}\right)$

$\left(\dfrac{1+a}{2},\dfrac{2+b}{2},\dfrac{1+c}{2}\right)=\left(\dfrac{3}{2},\dfrac{2}{2},\dfrac{-1}{2}\right)$

On comparing both sides, we get

$\dfrac{1+a}{2}=\dfrac{3}{2}$

$1+a=3$

$a=3-1$

$a=2$

Similarly,

$\dfrac{2+b}{2}=\dfrac{2}{2}$

$2+b=2$

$b=2-2$

$b=0$

And,

$\dfrac{1+c}{2}=\dfrac{-1}{2}$

$1+c=-1$

$c=-1-1$

$c=-2$

Hence, the coordinates of vertex S are (2,0,-2).