For convenience, we will assume that we have a unit cube (each side has length 1 unit) and we place it such that one corner of the cube is at the origin.

The unit vectors i, j, and k act in the x-, y-, and z-directions respectively. So in our diagram, since we have a unit cube,

OA = i
OC = j
OS = k

From the diagram, we see that to move from B to S, we need to go −1 unit in the x direction, −1 unit in the y-direction and up 1 unit in the z-direction. Since we have a unit cube, we can write:

BS = −i − j + k

and similarly:

CP = i − j + k

The scalar product for the vectors BS and CP is:

BS `*` CP = |BS| |CP| cos θ

where θ is the angle between BS and CP.

So the angle θ is given by

θ = arccos[ (BS • CP) ÷ ( |BS| |CP| ) ]

Now,

BS `*` CP

= (−i − j + k) `*` (i − j + k)

= 1 + 1 + 1

= 1

and

|BS| |CP|

`= sqrt((-1)^2 + (-1)^2 + 1^2)` ` × sqrt(1^2 + (-1)^2 + 1^2)`

`= (sqrt3)(sqrt3)`

= 3

So

`θ = arccos (1/3)`

θ = 70.5°

So the angle between the strings is `70.5°`. (In this situation we assume "angle" refers to the acute angle between the strings.)

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