## Thomas' Calculus 13th Edition

$\sqrt 3(-i+j+k)$
Formula to calculate the unit vector $\hat{\textbf{u}}$ is: $\hat{\textbf{u}}=\dfrac{u}{|u|}$ Given: $u=\dfrac{1}{2}i -\dfrac{1}{2}j-\dfrac{1}{2}k$; $|u|=\sqrt{(\dfrac{1}{2})^2+(\dfrac{-1}{2})^2+(\dfrac{-1}{2})^2}=\sqrt{\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}}=\dfrac{\sqrt {3}}{2}$ Thus, $\hat{\textbf{u}}=\dfrac{(\dfrac{1}{2}i -\dfrac{1}{2}j-\dfrac{1}{2}k)}{\dfrac{(\sqrt {3}}{2})}=\dfrac{1}{\sqrt 3}(i-j-k)$ and, $-3\hat{\text{u}}=\dfrac{-3}{\sqrt 3}(i-j-k)$ or, $=\sqrt 3(-i+j+k)$