A line makes angles `angle, beta, gamma and delta` with the diagonals of a cube. Show that `cos^(2)alpha+cos^(2)beta+cos^(2)gamma+cos^(2)delta=4//3`.

Explanation

Angle between two lines is geven by
`cos theta = (vecb_(1)cdot vecb_(2))/(abs(vecb_(1))abs(vecb_(2))) (vecb_(1) and vecb_(2) `are parallel vectors to given lines)
Four diagonals of cube are OE, AD, BG and ltbtgt OE is parallel to `a(i+ j + K)`
AD is parallel to `a(-hati +hatj + hatk)`
BG is parallel to `a(-hati -hatj + hatk)`
CF is parallel to `a(hati -hatj + hatk)`
Let the equation of given line is
`vecr = lambda (lhati+mhatj+nhatk)` (l, m, n, are direction cosines)
angle between `overset(rarr)(OE)` and given line
`cos alpha = (a(l+m+n))/(sqrt(3)a xx1) … (i) `
angle between `overset(rarr)(AD)` and given line
`cos beta = (a(-l+m+n))/(sqrt(3)a ) … (ii) `
angle between `overset(rarr)(BG)` and given line
`cos gamma = (a(-l-m+n))/(sqrt(3)a ) … (iii) `
angal between `overset (rarr)(CF)` and give line
`cos delta = (a(l+m+n))/(sqrt(3)a ) … (iv) `
Squaring and adding (i), (ii), (iii) and (iv)
`cos^(2) alpha + cos^(2)beta +cos^(2) gamma + cos^(2) delta = ((l+m+n)^(2)+(-l+m+n)^(2)+(-l-m+n)^(2)+(l-m+n)^(2))/3`
`= 4/3 (l^(2)+m^(2)+n^(2))=4/3`