## University Calculus: Early Transcendentals (3rd Edition)

$1(\dfrac{3}{5}i+\dfrac{4}{5}k)$
Here, $v=\dfrac{3}{5}i+\dfrac{4}{5}k$ and $|v|=\sqrt{(\dfrac{3}{5})^2+(\dfrac{4}{5})^2}=\sqrt {\dfrac{25}{25}}=1$ The unit vector $\hat{\textbf{v}}$ can be calculated as: $\hat{\textbf{v}}=\dfrac{v}{|v|}$ Now, $\hat{\textbf{v}}=\dfrac{(\dfrac{3}{5}i+\dfrac{4}{5}k)}{1}=(\dfrac{3}{5}i+\dfrac{4}{5}k)$ Thus, $v=|v|\hat{\textbf{v}}=1(\dfrac{3}{5}i+\dfrac{4}{5}k)$