Question

Let vector r₁ = a₁i + b₁j + c₁k

and vector r₂ = a₂i + b₂j + c₂k.

Then cos(θ) = ((a₁a₂ + b₁b₂ + c₁c₂)/(√(a₁² + b₁² + c₁²) √(a₂² + b₂² + c₂²)).

The angle between two diagonals of a unit cube is 

Answer 1

Answer is (c) tan^–1(2√2)

OP = i + j + k

BN = i + j – k

OP.OL = 1 + 1 

√3.√3 cos θ = 1 + 1

cos θ = 1/3

θ = cos^–1(1/3)

Hence, the angle between any two diagonals is

θ = cos^–1(1/3) = tan^–1(2√2)