Chapter 9 - Review - Exercises - Page 673: 8

$|v| = \sqrt 29$ $θ = -75.8°$

Length of a vector is defined as: $ v = ai + bj$ $|v| = \sqrt {a^2 + b^2}$ Direction is $\tan θ = \frac{b}{a}$ Given $v = 2i - 5j$ $|v| = \sqrt {(2)^2 + (-5)^2} = \sqrt 29$ $\tan θ = \frac{-5}{2} = -2.5$ $θ = -75.8°$