Three points are given A (-1; 3; -5) B (4; 2; -5) C (0; -2; -5). Calculate the cosine of the angle C.

Angle C is the angle between the CA and CB vectors.
1. Let’s define the coordinates of the CA vector:
CA = (- 1 – 0; 3 – (- 2); – 5 – (- 5)) = (- 1; 5; 0).
Let’s define the coordinates of the SV vector:
SV = (4 – 0; 2 – (- 2); – 5 – (- 5)) = (4; 4; 0).
2. Find the scalar product of the vectors CA and CB:
CA * SV = (-1) * 4 + 5 * 4 + 0 * 0 = – 4 + 20 + 0 = 16.
3. The scalar product of two vectors is equal to the product of their lengths and the cosine of the angle between them.
Let’s find the length of the CA vector:
| CA | = √ ((- 1) ^ 2 + 5 ^ 2 + 0 ^ 2) = √ (1 + 25 + 0) = √26.
Find the length of the SV vector:
| SV | = √ (4 ^ 2 + 4 ^ 2 + 0 ^ 2) = √ (16 + 16 + 0) = √32.
4. Thus:
| CA | * | CB | * cosC = CA * CB;
√26 * √32 * cosC = 16;
cosC = 16 / √832 = 16 / √16 * 4 * 13 = 16/4 * 2√13 = 16 / 8√13 = 2 / √13 = 2√13 / 13.
Answer: cosC = 2√13 / 13.