Prove that the smaller angle between any two diagonals of a cube is `cos^(-1)""(1)/(3)`.

Explanation

Let O be the origin of reference and A, B ,C vertices with position vectors, `veca,vecb and vecc` vertices A vector normal to plane ABC is
`veca xx vecb + vecb xx vecc xx veca and vec(OA0 =veca`
The angle between a line and a plane is equal to the normal to the plane. thus, if `theta` dentos the angle between the face and edge , then ,
`sin theta= ((vecb xxvecc +veccxxveca + veca xxvecb).veca)/(|vecbxxvecc + vecc xx veca +veca xxvecb |.|veca|)`
`([veca vecb vecc])/(|vecb xx vecc +vecc xxveca + veca xxvecb|.|veca|)`
Now,
`[veca vecb vecc]^(2)= |{:(veca.veca,veca.vecb,veca.vecc),(vecb.veca,vecb.vecb,vecb.vecc),(vecc.veca,vecc.vecb,vecc.vecc):}|`
`=k^6|{:(1,,"cos"60^@,,cos60^@),("cos"60^@,,1,,cos60^@),("cos"60^@,,"cos"60^@,,1):}|`
( where k is the length of the side of the tetrahedron )
` K^(6) (3/4 - 1/8 - 1/8) = 1/2 k^(6)`
Also `(vecb xx vecc + vecc xx veca + veca xx vecb)` is twice the area of triangle ABC, which is equilateral with each side k so that is `sqrt3/2 k^(2)` hence.
`sin theta(k^(3)/sqrt2)/(sqrt3/2 k^2.k) = 2/sqrt6 Rightarrow cos theta = 1/sqrt3`