## Thomas' Calculus 13th Edition

$\lt \dfrac{10}{\sqrt{35}},\dfrac{-2}{\sqrt{35}},\dfrac{-6}{\sqrt{35}} \gt$
The unit vector $\hat{\textbf{u}}$ can be calculated as: $\hat{\textbf{u}}=\dfrac{v}{|v|}$ Now, $v=\lt 5,-1,-3 \gt$ and $|v|=\sqrt{(5)^2+(-1)^2+(-3)^2}=\sqrt {35}$ Now, $2 \hat{\textbf{u}}=\dfrac{v}{|v|}=2 [\dfrac{\lt 5,-1,-3 \gt}{\sqrt {35}}]= \lt \dfrac{10}{\sqrt{35}},\dfrac{-2}{\sqrt{35}},\dfrac{-6}{\sqrt{35}} \gt$