Question

Show that the point $(5,-1,1)(7,-4,7),(1,-6,10)(-1,-3,4)$ are the vertices of a rhombus.

Answer 1

$A(5,-1,1)$

$B(7,-4,7)$

$C(1,-6,10)$

$D(-1,-3,4)$

Vertices

$AB=\sqrt{(7-5{)}^2+(-4+1{)}^2+(7-1{)}^2}=7$

$BC=\sqrt{(1-7{)}^2+(-6+4{)}^2+(10-7{)}^2}=7$

$CD=\sqrt{(-1-1{)}^2+(-3+6{)}^2+(4-10{)}^2}=7$

$DA=\sqrt{(-1-7{)}^2+(-3+4{)}^2+(4-7{)}^2}=7$

Diagonals

$AC=\sqrt{(1-5{)}^2+(-6+1{)}^2+(10-1{)}^2}=\sqrt{122}$

$BD=\sqrt{(-1-7{)}^2+(-3+4{)}^2+(4-7{)}^2}=\sqrt{74}$

Vertices are equal but the diagonals are not. Hence these are the vertices of rhombus.