## Thomas' Calculus 13th Edition

$1; \lt -3,0,-4 \gt$
As we know that the unit vector $\hat{\textbf{u}}$ is defined as: $\hat{\textbf{u}}=\dfrac{v}{|v|}$ Now, we have: $|v|=\sqrt{(\dfrac{3}{5})^2+0^2+(\dfrac{4}{5})^2}=1$ Thus, $-5 \hat{\textbf{u}}=\dfrac{v}{|v|}=-5 [\dfrac{\lt \dfrac{3}{5},0,\dfrac{4}{5} \gt}{1}]= \lt -3,0,-4 \gt$