Answer
$\dfrac{1}{\sqrt 3}i +\dfrac{1}{\sqrt 3}j +\dfrac{1}{\sqrt 3} k$
Work Step by Step
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:
$u=\dfrac{i+j+k}{\sqrt {(1)^2+(1)^2+(1)^2}} =\dfrac{i+j+k}{\sqrt {1+1+1}} \\=\dfrac{1}{\sqrt 3}i +\dfrac{1}{\sqrt 3}j +\dfrac{1}{\sqrt 3} k$