Answer
$${\bf{u}} \cdot {\bf{v}} = - 2,\,\,\,\,{\text{and}}\,\,\,\,\theta \approx {93.2^ \circ }$$
Work Step by Step
$$\eqalign{
& {\bf{u}} = \left\langle {4,3} \right\rangle {\text{ and }}{\bf{v}} = \left\langle {4, - 6} \right\rangle \cr
& {\text{find the dot product using the theorem 11}}{\text{.1 }}\left( {page\,\,783} \right) \cr
& {\bf{u}} \cdot {\bf{v}} = \left\langle {4,3} \right\rangle \cdot \left\langle {4, - 6} \right\rangle = \left( 4 \right)\left( 4 \right) + \left( 3 \right)\left( { - 6} \right) \cr
& {\bf{u}} \cdot {\bf{v}} = 16 - 18 \cr
& {\bf{u}} \cdot {\bf{v}} = - 2 \cr
& {\text{find the magnitude of }}{\bf{u}}{\text{ and }}{\bf{v}}\,\,\left( {see\,\,page\,\,\,\,776} \right) \cr
& \left| {\bf{u}} \right| = \left| {\left\langle {4,3} \right\rangle } \right| = \sqrt {{{\left( 4 \right)}^2} + {{\left( 3 \right)}^2}} = \sqrt {16 + 9} = 5 \cr
& \left| {\bf{v}} \right| = \left| {\left\langle {4, - 6} \right\rangle } \right| = \sqrt {{{\left( 4 \right)}^2} + {{\left( { - 6} \right)}^2}} = \sqrt {16 + 36} = \sqrt {52} \cr
& {\text{find the angle between the vectores using }}\cos \theta = \frac{{{\bf{u}} \cdot {\bf{v}}}}{{\left| {\bf{u}} \right|\left| {\bf{v}} \right|}}{\text{ then}} \cr
& \cos \theta = \frac{{{\bf{u}} \cdot {\bf{v}}}}{{\left| {\bf{u}} \right|\left| {\bf{v}} \right|}} = \frac{{ - 2}}{{\left( 5 \right)\left( {\sqrt {52} } \right)}} \cr
& {\text{simplifying}} \cr
& {\text{cos}}\theta = - \frac{2}{{5\left( {2\sqrt {13} } \right)}} \cr
& {\text{cos}}\theta = - \frac{1}{{5\sqrt {13} }} \cr
& {\text{solving for }}\theta \cr
& \theta = {\cos ^{ - 1}}\left( { - \frac{1}{{5\sqrt {13} }}} \right) \cr
& {\text{simplify by using a calculator}} \cr
& \theta \approx {93.2^ \circ } \cr} $$
