Respuesta :
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.
[tex]\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)[/tex]
[tex]\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)[/tex]
On comparing both sides, we get
[tex]\dfrac{1+a}{2}=\dfrac{3}{2}[/tex]
[tex]1+a=3[/tex]
[tex]a=3-1[/tex]
[tex]a=2[/tex]
Similarly,
[tex]\dfrac{2+b}{2}=\dfrac{2}{2}[/tex]
[tex]2+b=2[/tex]
[tex]b=2-2[/tex]
[tex]b=0[/tex]
And,
[tex]\dfrac{1+c}{2}=\dfrac{-1}{2}[/tex]
[tex]1+c=-1[/tex]
[tex]c=-1-1[/tex]
[tex]c=-2[/tex]
Therefore, the coordinates of vertex S are (2,0,-2).
