Points A (-1; 2; 2), B (4; 2; 2) are given. Find the length of the vector AB.

The length of an arbitrary vector A1A2, given by the coordinates of the beginning of A1 (x1; y1; z1) and the end of A2 (x2; y2; z2), can be calculated as

A1A2 = ((x2 – x1) ^ 2 + (y2 – y1) ^ 2 + (z2 – z1) ^ 2) ^ 0.5.

Thus, the length of the vector AB, given by the coordinates A (-1; 2; 2) and B (4; 2; 2), is equal to

AB = ((4 – (-1)) ^ 2 + (2 – 2) ^ 2 + (2 – 2) ^ 2) ^ 0.5;

AB = (5 ^ 2 + 0 ^ 2 + 0 ^ 2) ^ 0.5;

AB = 5.

Answer: The length of vector AB is 5 linear units.