Show that the points `A(3,3,3,),\ B(0,6,3),\ C(1,7,7)a n d\ D(4,4,7)` are the vertices of a square.

Explanation

Given points are `A(3,3,3,),\ B(0,6,3),\ C(1,7,7)a n d\ D(4,4,7)`
we have to Show that the points `A,B,C,D`
are the vertices of a square.
So we have to show that `AB=BC=CD=DA` and also diagonal to be equal i.e. `AC=BD`
Now `AB=sqrt((3-0)^2+(3-6)^2+(3-3)^2)`
`=sqrt((3)^2+(-3)^2+(0)^2)`
`=sqrt(9+9)`
`=sqrt(18)`
`=3sqrt(2)`
`BC=sqrt((0-1)^2+(6-7)^2+(3-7)^2)`
`=sqrt((-1)^2+(-1)^2+(-4)^2)`
`=sqrt(1+1+16)`
`=sqrt(18)`
`=3sqrt(2)`
`CD=sqrt((1-4)^2+(7-4)^2+(7-7)^2)`
`=sqrt((-3)^2+(3)^2+(0)^2)`
`=sqrt(9+9)`
`=sqrt(18)`
`=3sqrt(2)`
`DA=sqrt((4-3)^2+(4-3)^2+(7-3)^2)`
`=sqrt((1)^2+(1)^2+(4)^2)`
`=sqrt(1+1+16)`
`=sqrt(18)`
`=3sqrt(2)`
We can clearly see that `AB=BC=CD=DA`
Now we have to show two diagonals to be equal.
`AC=sqrt((3-1)^2+(3-7)^2+(3-7)^2)`
`=sqrt((2)^2+(-4)^2+(-4)^2)`
`=sqrt(4+16+16)`
`=sqrt(36)`
`=6`
`BD=sqrt((0-4)^2+(6-4)^2+(3-7)^2)`
`=sqrt((-4)^2+(2)^2+(-4)^2)`
`=sqrt(4+16+16)`
`=sqrt(36)`
`=6`
Since `AB=BC=CD=DA`and `AC=BD`
Therefore `A ,B ,C, D` are 4 vertices of a square .