## Thomas' Calculus 13th Edition

$11(\dfrac{9}{11}i-\dfrac{-2}{11}j+\dfrac{6}{11}k)$
Formula to find the unit vector $\hat{\textbf{v}}$ is: $\hat{\textbf{v}}=\dfrac{v}{|v|}$ Given: $v=9i-2j+6k$ and $|v|=\sqrt{(9)^2+(-2)^2+(6)^2}=\sqrt{81+4+36}=\sqrt {121}=11$ Thus, $\hat{\textbf{v}}=\dfrac{9i-2j+6k}{11}=(\dfrac{9}{11},\dfrac{-2}{11},\dfrac{6}{11})$ and $v=|v|\hat{\textbf{v}}=11(\dfrac{9}{11}i-\dfrac{-2}{11}j+\dfrac{6}{11}k)$