For this case we have the following vector:
v = (- 2, -1).
Using product point we have:
v.v = (-2, -1). (- 2, -1)
v ^ 2 = (-2) * (- 2) + (-1) * (- 1)
v ^ 2 = 4 + 1
v ^ 2 = 5
Then, clearing the module we have:
v = root (5)
Answer:
The module of the vector is given by:
v = root (5)
C. square root of 5
