Answer
$v=(0, \sqrt 6, \frac{\sqrt 6}{2}, -\frac{\sqrt 6}{2})$
Work Step by Step
The magnitude of vector, $u$ is given by $\sqrt {(0^{2}+(2)^{2}+(1)^{2}+(-1)^{2}}=\sqrt {6}$
Hence a unit vector in the same direction as $u$ is given by $\frac{1}{\sqrt {6}}\times(02,1,-1)$
$=(0,\frac{2}{\sqrt {6}}, \frac{1}{\sqrt {6}}, - \frac{1}{\sqrt {6}})$
Vector, $v$ has length 3, so must be equal to $3\times(0,\frac{2}{\sqrt {6}}, \frac{1}{\sqrt {6}}, - \frac{1}{\sqrt {6}})$
$=(0, \frac{6}{\sqrt {6}}, \frac{3}{\sqrt {6}} - \frac{3}{\sqrt {6}})$
Therefore $v=(0, \sqrt 6, \frac{\sqrt 6}{2}, -\frac{\sqrt 6}{2})$
