Solve this vector question?

Two space diagonals of a cube are shown in bold in the diagram.("sorry i could not paste the diagram")
The diagonals of a square meet at right,angles. It might be supposed that two space
diagonals of a cube would also meet at right angles.
Surprisingly, this is not the case. What is the angle between two space diagonals of a cube?
Show your working please, thanks
hint: use scalar product to find the answer
a.b=lallbl cos theta

Use a unit cube with a vertex at the origin O. Choose A(1, 1, 1), B(0, 1, 1) and C(1, 0, 0) then space diagonals OA = a = [1, 1, 1] and CB = b = [-1, 1, 1] then
a . b = -1 + 1 + 1 = 1 which is not zero and so a and b are not perpendicular

Calculate the angle 胃 between the space diagonals of a cube.

Take a cube with edge of length one and vertices at O(0,0,0); A(1,1,1); B(1,1,0); C(0,1,1), etc.

The angle between any two of the four body diagonals will be the same. Let's look at the diagonals OA and BC.

OA = <1-0, 1-0, 1-0> = <1, 1, 1>
BC = <1-0, 1-0, 0-1> = <1, 1, -1>

Calculate the magnitude of the vectors.

|| OA || = 鈭歔1虏 + 1虏 + 1虏] = 鈭?
|| BC || = 鈭歔1虏 + 1虏 + (-1)虏] = 鈭?

Take the dot product.

OA 鈥?BC = <1, 1, 1> 鈥?<1, 1, -1> = 1 + 1 - 1 = 1

The dot product can also be expressed another way.

OA 鈥?BC = || OA || || BC || cos胃
1 = (鈭?)(鈭?) cos胃
1 = 3cos胃
cos胃 = 1/3
胃 = arccos(1/3) 鈮?70.528779掳

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