text: ## Algebra and Trigonometry 10th Edition
  
  $\lt \dfrac {18}{\sqrt {29}}, \dfrac{ 45}{\sqrt {29}} \gt$
  Since $v$ has the same direction as $u$ and since it is a multiple of a unit vector $n$, we can write it as: $\overrightarrow{v} = < 2 n, 5n>$ To find the unit vector, we must first find the magnitude: $\sqrt {((2n)^{2} + (5n)^{2}} = 3$ or, $144 \ n^2+25 \ n^2 =9$ This implies that $n^2=\dfrac{ 81}{29} \implies n=\dfrac{9}{\sqrt {29}}$ Now, $\overrightarrow{v} = < 2 (\dfrac{9}{\sqrt {29}}), 5 (\dfrac{9}{\sqrt {29}})> = \lt \dfrac {18}{\sqrt {29}}, \dfrac{ 45}{\sqrt {29}} \gt$