Angles of a Triangle via the Dot Product:

A(3, 4, 2), B(-2, 5, 4), C(2, -1, -3).

Let the angles be q _A, q _B, and q _C, respectively.

To determine q _A we need the vectors AB and AC.

AB = (-2 -3)i + (5 - 4)j + (4 - 2)k = -5i + j + 2k.

AC = (2 -3)i + (-1 - 4)j + (-3 - 2)k = -i - 5j - 5k.

 . This becomes

Solving for the angle, we get

 Similarly, for q _B we need the vectors BA and BC.

BA = -AB = 5i - j - 2k.

BC = (2 -(-2))i + (-1 - 5)j + (-3 - 4)k = 4i - 6j - 7k.

 . This becomes

Solving for the angle, we get

 For q _C we need the vectors CA and CB.

CA = -AC = i + 5 j + 5k.

CB =-BC = -4i + 6j + 7k.

 . This becomes

Solving for the angle, we get

 We see that the sum of these three angles is approximately 180° .