Question

Find the angle between two diagonals of a cube.

Solution

Take O, a corner of cube OBLCMANP, as origin and OA, OB, OC, the three edges through it as the axes.

Let OA = OB = OC = a, then the co-ordinates of O, A, B, C are (0, 0, 0), (a, 0, 0), (0, a, 0), (0, 0, a) respectively ; those of P, L, M, N are (a, a, a), (0, a, a), (a, 0, a), (a, a, 0) respectively.
The four diagonals are OP, AL, BM, CN. Direction cosines of OP are proportional to a – 0, a – 0, a – 0, i.e., a, a, a, i.e., 1, 1, 1.
Direction-cosines of AL are proportional to 0 – a, a – 0, a – 0    i.e., –a, a, a, i.e., – 1, 1, 1.
Direction-cosines of BM are proportional to a – 0, 0 – a, a – 0. i.e.. a – a, a    i.e., 1, – 1, 1.
Direction-cosines of CN are proportional to a – 0, a – 0, 0 – a    i.e., a, a, – a    i.e., 1, 1, – 1.
Let θ be the angle between AL and BM
$therefore space space space space space space cos space straight theta space equals space open vertical bar fraction numerator left parenthesis negative 1 right parenthesis thin space left parenthesis 1 right parenthesis space plus space left parenthesis 1 right parenthesis thin space left parenthesis negative 1 right parenthesis space plus space left parenthesis 1 right parenthesis thin space left parenthesis 1 right parenthesis over denominator square root of 1 plus 1 plus 1 end root space square root of 1 plus 1 plus 1 end root end fraction close vertical bar space equals space open vertical bar fraction numerator negative 1 minus 1 plus 1 over denominator square root of 3. square root of 3 end fraction close vertical bar space equals space 1 third therefore space space space space straight theta space equals space cos to the power of negative 1 end exponent 1 third$
Similarly the angle between other two diagonals is also cos $1 third$ .

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Three Dimensional Geometry

Find the angle between two diagonals of a cube.

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