Answer
$$\left\langle {0,\frac{{10}}{{\sqrt 2 }},\frac{{10}}{{\sqrt 2 }}} \right\rangle $$
Work Step by Step
$$\eqalign{
& \left\| {\bf{v}} \right\| = 10,{\text{ }}{\bf{u}} = \left\langle {0,3,3} \right\rangle \cr
& {\text{Calculate }}\left\| {\bf{u}} \right\| \cr
& \left\| {\bf{u}} \right\| = \sqrt {{{\left( 0 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 3 \right)}^2}} \cr
& \left\| {\bf{u}} \right\| = 3\sqrt 2 \cr
& {\text{A unit vector in the direction of }}{\bf{u}}{\text{ is}} \cr
& \frac{{\bf{u}}}{{\left\| {\bf{u}} \right\|}} = \frac{{\left\langle {0,3,3} \right\rangle }}{{3\sqrt 2 }} = \left\langle {0,\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right\rangle \cr
& \frac{{\bf{u}}}{{\left\| {\bf{u}} \right\|}} = \left\langle {0,\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right\rangle \cr
& {\text{A vector with magnitude 10 in the direction of }}{\bf{u}}{\text{ is}} \cr
& = 10\left\langle {0,\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right\rangle \cr
& = \left\langle {0,\frac{{10}}{{\sqrt 2 }},\frac{{10}}{{\sqrt 2 }}} \right\rangle \cr} $$
