Given three points A (0; 0; 1), B (1; -2; 2), C (3; 0; 0) Find the area A of triangle ABC.

Angle A can be calculated using the scalar product of vectors AB (1; -2; 1) and AC (3; 0; -1).

Based on the geometric and algebraic interpretation of the definition of the concept of “scalar product of two vectors” we get

cos A = (x (AB) * x (AC) + y (AB) * y (AC) + z (AB) * z (AC)) / (x (AB) ^ 2 + y (AB) ^ 2 + z (AB) ^ 2) * (x (AC) ^ 2 + y (AC) ^ 2 + z (AC) ^ 2) = (1 * 3 + (-2) * 0 + 1 * (-1)) / (1 ^ 2 + (-2) ^ 2 + 1 ^ 2) * (3 ^ 2 + 0 ^ 2 + (-1) ^ 2) = 2/60 = 1/30.
Thus, the angle A = arccos 1/30 = 88 ° approximately.