Question

Show that the points A(3, 3, 3), B(0, 6, 3), C(1, 7, 7) and D(4, 4, 7)

Answer 1

AB=$\sqrt{(0-3{)}^2+(6-3{)}^2+(3-3{)}^2}$

=$\sqrt{(-3{)}^2+(3{)}^2+(0{)}^2}=\sqrt{9+9}$

=$3\sqrt{2}$ units

BC=$\sqrt{(1-0{)}^2+(7-6{)}^2+(7-3{)}^2}$

=$\sqrt{(1{)}^2+(1{)}^2+(4{)}^2}$

=$\sqrt{1+1+16}=3\sqrt{2}$ units

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

=$\sqrt{(-2{)}^2+(4{)}^2+(4{)}^2}$

=$\sqrt{4+16+16}=\sqrt{36}=6$ units

BD =$\sqrt{(4-0{)}^2+(4-6{)}^2+(7-3{)}^2}$

= $\sqrt{(4{)}^2+(-2{)}^2+(4{)}^2}$

=$\sqrt{16+4+16}=\sqrt{36}=6$ units

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

=$\sqrt{(3{)}^2+(-3{)}^2+(0{)}^2}$

=$\sqrt{9+9}$

=3$\sqrt{2}$ units

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

=${\sqrt{\left(1\right)+(1{)}^2+(4)}}^2=\sqrt{1+1+16}$

=$3\sqrt{2}$ units

Since,

AB = BC = CD = DA

And AC = BD

Since, all sides and diagnols of quadrilateral ABCD all equal.