Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\dfrac{2}{\sqrt 6}i -\dfrac{1}{\sqrt 6}j +\dfrac{1}{\sqrt 6} k$
The magnitude of any vector (let us say $v$) can be determined using the formula $||v||=\sqrt{p^2+q^2+r^2}$ and the unit vector $u$ in the same direction as $v$ can be found as: $u=\dfrac{v}{||v||}$ Therefore, the unit vector $u$ in the same direction as $v$ is: $u=\dfrac{2i-j+k}{\sqrt {(2)^2+(-1)^2+(1)^2}} =\dfrac{2i-j+k}{\sqrt {4+1+1}} \\=\dfrac{2}{\sqrt 6}i -\dfrac{1}{\sqrt 6}j +\dfrac{1}{\sqrt 6} k$