Question

The angle between two diagonals of a cube with edges of unit length

Answer 1

Given: The edge of a cube is 1 unit .

To find: The angle between two diagonals of a cube ?

Solution:

• Let the cube be OABCDEFG, The vertices will be:

                    O(0,0,0), A(1,0,0), B(1,1,0), C(0,1,0), D(0,1,1), E(0,0,1), F(1,0,1), G(1,1,1)

• Now the diagonals are:

                    OG, CF, AD and BE.

• For AB, we have:

                    AB = (x2-x2)i + (y2-y1)j + (z2-z1)k

• So lets consider two diagonals, OG and AD, we get:

                    OG  = (1−0)i + (1−0)j + (1−0)k

                    OG = i + j + k

                     AD  = (0−1)i +(1−0)j+(1−0)k

                    AD = -i + j + k

• Then :

                    |OG| = √1²+1²+1² = √3

                    |AD| = √(-1)²+1²+1² = √3

• Now

                    OG.AD =  i + j + k  . -i + j + k = -1^2 + 1^2 + 1^2 = 1

• Now the angle between two vectors   a  ,  b  is:

                    theta = cos^-1 (a.b / |a||b| )

• So putting the values in it, we get:

                    theta = cos^-1 (1 / √3√3 )

                    theta = cos^-1(1/3)

Answer:

           So the angle between two diagonals is cos^-1(1/3).