Given points A (0; 4), B (4; 2), C (2; -2), D (-2; 0) Find the coordinates and length of the vector a = AB + 3AD-1 / 2CA

1. Given:

A (0; 4);
B (4; 2);
C (2; -2);
D (-2; 0);
a = AB + 3AD – 1/2 * CA.
2. Find the coordinates of the vectors:

AB = (4; 2) – (0; 4) = (4 – 0; 2 – 4) = (4; -2);
AD = (-2; 0) – (0; 4) = (-2 – 0; 0 – 4) = (-2; -4);
CA = (0; 4) – (2; -2) = (0 – 2; 4 + 2) = (-2; 6);
a = AB + 3 * AD – 1/2 * CA;
a = (4; -2) + 3 (-2; -4) – 1/2 (-2; 6) = (4; -2) + (-6; -12) + (1; -3) = (-1; -17).
3. Length of vector a:

| a | ^ 2 = (-1) ^ 2 + (-17) ^ 2 = 1 + 289 = 290;
| a | = √290.
Answer:

1) coordinates of the vector a: (-1; -17);
2) its length: √290.