Answer
$${\bf{u}} \cdot {\bf{v}} = - 4,\,\,\,\,{\text{and}}\,\,\,\,\theta = \pi $$
Work Step by Step
$$\eqalign{
& {\bf{u}} = \left\langle {\sqrt 2 ,\sqrt 2 } \right\rangle {\text{ and }}{\bf{v}} = \left\langle { - \sqrt 2 , - \sqrt 2 } \right\rangle \cr
& {\text{find the dot product using the theorem 11}}{\text{.1 }}\left( {page\,\,783} \right) \cr
& {\bf{u}} \cdot {\bf{v}} = \left\langle {\sqrt 2 ,\sqrt 2 } \right\rangle \cdot \left\langle { - \sqrt 2 , - \sqrt 2 } \right\rangle = \left( {\sqrt 2 } \right)\left( { - \sqrt 2 } \right) + \left( {\sqrt 2 } \right)\left( { - \sqrt 2 } \right) \cr
& {\bf{u}} \cdot {\bf{v}} = - {\left( {\sqrt 2 } \right)^2} - {\left( {\sqrt 2 } \right)^2} \cr
& {\bf{u}} \cdot {\bf{v}} = - 4 \cr
& {\text{find the magnitude of }}{\bf{u}}{\text{ and }}{\bf{v}}\,\,\left( {see\,\,page\,\,\,\,776} \right) \cr
& \left| {\bf{u}} \right| = \left| {\left\langle {\sqrt 2 ,\sqrt 2 } \right\rangle } \right| = \sqrt {{{\left( {\sqrt 2 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} = \sqrt {2 + 2} = 2 \cr
& \left| {\bf{v}} \right| = \left| {\left\langle { - \sqrt 2 , - \sqrt 2 } \right\rangle } \right| = \sqrt {{{\left( { - \sqrt 2 } \right)}^2} + {{\left( { - \sqrt 2 } \right)}^2}} = \sqrt {2 + 2} = 2 \cr
& {\text{find the angle between the vectores using }}\cos \theta = \frac{{{\bf{u}} \cdot {\bf{v}}}}{{\left| {\bf{u}} \right|\left| {\bf{v}} \right|}}{\text{ then}} \cr
& \cos \theta = \frac{{{\bf{u}} \cdot {\bf{v}}}}{{\left| {\bf{u}} \right|\left| {\bf{v}} \right|}} = \frac{{ - 4}}{{\left( 2 \right)\left( 2 \right)}} \cr
& {\text{simplifying}} \cr
& {\text{cos}}\theta = - 1 \cr
& {\text{solving for }}\theta \cr
& \theta = {\cos ^{ - 1}}\left( { - 1} \right) \cr
& {\text{simplify}} \cr
& \theta = \pi \cr
& \cr} $$
