Use the formula for magnitude of vector, which is [tex]m = \sqrt{a^2 + b^2}[/tex].
[tex]m = \sqrt{(-10)^2 + 13^2} = 16.401[/tex] units
In terms of direction, you can see that it is in the second quadrant because of a negative x-value and positive y-value. If you are looking for the angle of the vector, you need to use the formula,Ф = [tex]arctan(y/x)[/tex]. But because the x, value is negative, the rule is that we add π to this value.
Plug in the values:
arctan(13/-10) + π = 127.56° or 2.226 radians
