Answer
$$\eqalign{
& \left( {\bf{a}} \right)1.67 \cr
& \left( {\bf{b}} \right)96.05^\circ \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Let }}{\bf{u}} = \left\langle {1,0, - 3} \right\rangle ,{\text{ }}{\bf{v}} = \left\langle {2, - 2,1} \right\rangle \cr
& {\bf{u}} \cdot {\bf{v}} = \left\langle {1,0, - 3} \right\rangle \cdot \left\langle {2, - 2,1} \right\rangle \cr
& {\bf{u}} \cdot {\bf{v}} = 2 + 0 - 3 \cr
& {\bf{u}} \cdot {\bf{v}} = - 1 \cr
& \left\| {\bf{u}} \right\| = \left\| {\left\langle {1,0, - 3} \right\rangle } \right\| = \sqrt {1 + 0 + 9} = \sqrt {10} \cr
& \left\| {\bf{v}} \right\| = \left\| {\left\langle {2, - 2,1} \right\rangle } \right\| = \sqrt {4 + 4 + 1} = 3 \cr
& {\text{Find the Angle Between Two Vectors}} \cr
& \cos \theta = \frac{{{\bf{u}} \cdot {\bf{v}}}}{{\left\| {\bf{u}} \right\|\left\| {\bf{v}} \right\|}} \cr
& \cos \theta = \frac{{ - 1}}{{3\sqrt {10} }} \cr
& \left( {\bf{a}} \right){\text{ In radians}} \cr
& \theta = {\cos ^{ - 1}}\left( {\frac{{ - 1}}{{3\sqrt {10} }}} \right) \cr
& \theta = 1.67 \cr
& \left( {\bf{b}} \right){\text{ In degrees}} \cr
& \theta = 96.05^\circ \cr} $$
