Answer
$$\left\langle { - \sqrt {14} ,\frac{{3\sqrt {14} }}{2},\frac{{\sqrt {14} }}{2}} \right\rangle $$
Work Step by Step
$$\eqalign{
& \left\| {\bf{v}} \right\| = 7,{\text{ }}{\bf{u}} = \left\langle { - 4,6,2} \right\rangle \cr
& {\text{Calculate }}\left\| {\bf{u}} \right\| \cr
& \left\| {\bf{u}} \right\| = \sqrt {{{\left( { - 4} \right)}^2} + {{\left( 6 \right)}^2} + {{\left( 2 \right)}^2}} \cr
& \left\| {\bf{u}} \right\| = \sqrt {56} \cr
& \left\| {\bf{u}} \right\| = 2\sqrt {14} \cr
& {\text{A unit vector in the direction of }}{\bf{u}}{\text{ is}} \cr
& \frac{{\bf{u}}}{{\left\| {\bf{u}} \right\|}} = \frac{{\left\langle { - 4,6,2} \right\rangle }}{{2\sqrt {14} }} = \left\langle { - \frac{{\sqrt {14} }}{7},\frac{{3\sqrt {14} }}{{14}},\frac{{\sqrt {14} }}{{14}}} \right\rangle \cr
& {\text{A vector with magnitude 7 in the direction of }}{\bf{u}}{\text{ is}} \cr
& = 7\left\langle { - \frac{{\sqrt {14} }}{7},\frac{{3\sqrt {14} }}{{14}},\frac{{\sqrt {14} }}{{14}}} \right\rangle \cr
& = \left\langle { - \sqrt {14} ,\frac{{3\sqrt {14} }}{2},\frac{{\sqrt {14} }}{2}} \right\rangle \cr} $$
