To prove: Points A, B, C, D form square.
Formula: The distance between two points (x1,y1,z1) and (x2,y2,z2) is given by
D = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\)
Here,
(x1,y1,z1)= (1, 1, 1)
(x2,y2,z2)= (-2, 4, 1)
(x3,y3,z3)= (-1, 5, 5)
(x4,y4,z4)= (2, 2, 5)
Length AB = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\)
Length BC = \(\sqrt{(x_3-x_2)^2+(y_3-y_2)^2+(z_3-z_2)^2}\)
Length AD = \(\sqrt{(x_4-x_1)^2+(y_4-y_1)^2+(z_4-z_1)^2}\)
Here, AB = BC = CD = AD
Also, AC = BD
This means all the sides are the same and diagonals are also equal.
Hence vertices A, B, C, D form a square.