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asked ago in Three-dimensional geometry by (20.3k points)

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 

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answered ago by (21.7k points)
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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)

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