Answer
$1(\dfrac{1}{\sqrt 3}i +\dfrac{1}{\sqrt 3}j +\dfrac{1}{\sqrt 3}k)$
Work Step by Step
Here, $v=\dfrac{1}{\sqrt 3}i +\dfrac{1}{\sqrt 3}j +\dfrac{1}{\sqrt 3}k$
and $|v|=\sqrt{(\dfrac{1}{\sqrt 3})^2+(\dfrac{1}{\sqrt 3})^2+(\dfrac{1}{\sqrt 3})^2}=\sqrt {1}=1$
The unit vector $\hat{\textbf{v}}$ can be calculated as: $\hat{\textbf{v}}=\dfrac{v}{|v|}$
Now, $\hat{\textbf{v}}=\dfrac{\dfrac{1}{\sqrt 3}i +\dfrac{1}{\sqrt 3}j +\dfrac{1}{\sqrt 3}k}{1}=(\dfrac{1}{\sqrt 3}i +\dfrac{1}{\sqrt 3}j +\dfrac{1}{\sqrt 3}k)$
Thus, $v=|v|\hat{\textbf{v}}=1(\dfrac{1}{\sqrt 3}i +\dfrac{1}{\sqrt 3}j +\dfrac{1}{\sqrt 3}k)$
