Answer
$${\bf{u}} \cdot {\bf{v}} = 6,\,\,\,\,{\text{and}}\,\,\,\,\theta \approx {80^ \circ }$$
Work Step by Step
$$\eqalign{
& {\bf{u}} = \left\langle {1, - 4, - 6} \right\rangle {\text{ and }}{\bf{v}} = \left\langle {2, - 4,2} \right\rangle \cr
& {\text{find the dot product using the theorem 11}}{\text{.1 }}\left( {page\,\,783} \right) \cr
& {\bf{u}} \cdot {\bf{v}} = \left\langle {1, - 4, - 6} \right\rangle \cdot \left\langle {2, - 4,2} \right\rangle = \left( 1 \right)\left( 2 \right) + \left( { - 4} \right)\left( { - 4} \right) + \left( { - 6} \right)\left( 2 \right) \cr
& {\bf{u}} \cdot {\bf{v}} = 2 + 16 - 12 \cr
& {\bf{u}} \cdot {\bf{v}} = 6 \cr
& {\text{find the magnitude of }}{\bf{u}}{\text{ and }}{\bf{v}}\,\,\left( {see\,\,page\,\,\,\,776} \right) \cr
& \left| {\bf{u}} \right| = \left| {\left\langle {1, - 4, - 6} \right\rangle } \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( { - 6} \right)}^2}} = \sqrt {1 + 16 + 36} = \sqrt {53} \cr
& \left| {\bf{v}} \right| = \left| {\left\langle {2, - 4,2} \right\rangle } \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 2 \right)}^2}} = \sqrt {4 + 16 + 4} = \sqrt {24} = 2\sqrt 6 \cr
& {\text{find the angle between the vectores using }}\cos \theta = \frac{{{\bf{u}} \cdot {\bf{v}}}}{{\left| {\bf{u}} \right|\left| {\bf{v}} \right|}}{\text{ then}} \cr
& \cos \theta = \frac{{{\bf{u}} \cdot {\bf{v}}}}{{\left| {\bf{u}} \right|\left| {\bf{v}} \right|}} = \frac{6}{{\left( {\sqrt {53} } \right)\left( {2\sqrt 6 } \right)}} \cr
& {\text{simplifying}} \cr
& \cos \theta = \frac{3}{{\left( {\sqrt {53} } \right)\left( {\sqrt 6 } \right)}} \cr
& {\text{solving for }}\theta \cr
& \theta = {\cos ^{ - 1}}\left( {\frac{3}{{\left( {\sqrt {53} } \right)\left( {\sqrt 6 } \right)}}} \right) \cr
& {\text{simplify by using a calculator}} \cr
& \theta \approx {80^ \circ } \cr} $$
