Given the vertices of the triangle A, B, C. Find the cosine of the angle BAC A (3; 3; –1); B (1; –5; 2); C (4; 4; 1).

Let’s find the lengths of the sides of the triangle.

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

| AB | = ((1 – 3) ^ 2 + (-5 – 3) ^ 2 + (2 + 1) ^ 2) ^ (1/2) = (4 + 64 + 9) ^ (1/2) = 77 ^ (1/2);

| AC | = ((4 – 3) ^ 2 + (4 – 3) ^ 2 + (1 + 1) ^ 2) ^ (1/2) = 6 ^ (1/2);

| BC | = ((4 – 1) ^ 2 + (4 + 5) ^ 2 + (1 – 2) ^ 2) ^ (1/2) = (9 + 81 + 1) ^ (1/2) = 91 ^ ( 1/2);

Let us write the cosine theorem for the side AC:

AC ^ 2 = AB ^ 2 + BC ^ 2 – 2 * AB * BC * cos BAC;

6 = 77 + 91 – 2 * (77 * 91) ^ (1/2) * cos BAC;

6 = 168 – 2 * 7 * 143 ^ (1/2) * cos BAC;

cos BAC = 162 / (14 * 143 ^ (1/2));

cos BAC = 0.968.