## Precalculus (10th Edition)

$\frac{1}{\sqrt{3}}i+\frac{1}{\sqrt{3}}j+\frac{1}{\sqrt{3}}k.$
If $v=ai+bj+ck$ and $||v||=\sqrt{a^2+b^2+c^2}$, then the unit vector $u$ in the same direction as $v$ is $u=\frac{v}{||v||}$. The magnitude of $v$ is $||v||=\sqrt{1^2+1^2+1^2}\\||v||=\sqrt{1+1+1}=\sqrt{3}.$ Thus the unit vector is: $u=\frac{i+j+k}{\sqrt{3}}\\u=\frac{1}{\sqrt{3}}i+\frac{1}{\sqrt{3}}j+\frac{1}{\sqrt{3}}k.$