Show that the angle between two diagonals of a cube is `cos^(-1)sqrt(1/3)dot`

Explanation

Take origin O as one vertex of the cube and OA, OB and OC as the positiv directions of the X-axis, the Y-axis and the Z-axis respectively. Here, the side of the cube are
`OA = OB = OC = a….` (Say)
`therefore` the coordinates of all the vertices of the cube will be

`{:(O-=(0,,0,,0)," "A-=(a,,0,,0)),(B-=(0,,a,,0)," "C-=(0,,0,,a)),(N-=(a,,a,,0)," "L-=(0,,a,,a)),(M-=(a,,0,,a)," "P-=(a,,a,,a)):}`
Here the four diagonals are AL, BM, CN and OP.
Consider the diagonals OP and AL.
The direction ratios of OP are `a -0, a-0, a-0,` i.e., a,a,a
The direction ratios of AL are `0-a, a-0, a-0`, i.e., `-a, a,a`
Let `theta` be the angle between the diagonals OP and AL.
Then `cos theta=|(a(-a)+a(a)+a(a))/(sqrt(a^(2)+a^(2)+a^(2))*sqrt((-a)^(2)+a^(2)+a^(2)))|`
` = |(-a^(2)+a^(2)+a^(2))/((sqrt(3)a)(sqrt(3)a))|=(1)/(3) " "therefore theta = cos^(-1)((1)/(3)).`