#### Answer

$${\bf{u}} \cdot {\bf{v}} = 2,\,\,\,\,{\text{and}}\,\,\,\,\theta \approx {87.2^ \circ }$$

#### Work Step by Step

$$\eqalign{
& {\bf{u}} = \left\langle { - 10,0,4} \right\rangle {\text{ and }}{\bf{v}} = \left\langle {1,2,3} \right\rangle \cr
& {\text{find the dot product using the theorem 11}}{\text{.1 }}\left( {page\,\,783} \right) \cr
& {\bf{u}} \cdot {\bf{v}} = \left\langle { - 10,0,4} \right\rangle \cdot \left\langle {1,2,3} \right\rangle = \left( { - 10} \right)\left( 1 \right) + \left( 0 \right)\left( 2 \right) + \left( 4 \right)\left( 3 \right) \cr
& {\bf{u}} \cdot {\bf{v}} = - 10 + 0 + 12 \cr
& {\bf{u}} \cdot {\bf{v}} = 2 \cr
& {\text{find the magnitude of }}{\bf{u}}{\text{ and }}{\bf{v}}\,\,\left( {see\,\,page\,\,\,\,776} \right) \cr
& \left| {\bf{u}} \right| = \left| {\left\langle { - 10,0,4} \right\rangle } \right| = \sqrt {{{\left( { - 10} \right)}^2} + {{\left( 0 \right)}^2} + {{\left( 4 \right)}^2}} = \sqrt {100 + 16} = \sqrt {116} \cr
& \left| {\bf{v}} \right| = \left| {\left\langle {1,2,3} \right\rangle } \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 3 \right)}^2}} = \sqrt {1 + 4 + 9} = \sqrt {14} \cr
& {\text{find the angle between the vectores using }}\cos \theta = \frac{{{\bf{u}} \cdot {\bf{v}}}}{{\left| {\bf{u}} \right|\left| {\bf{v}} \right|}}{\text{ then}} \cr
& \cos \theta = \frac{{{\bf{u}} \cdot {\bf{v}}}}{{\left| {\bf{u}} \right|\left| {\bf{v}} \right|}} = \frac{2}{{\left( {\sqrt {116} } \right)\left( {\sqrt {14} } \right)}} \cr
& {\text{simplifying}} \cr
& \cos \theta = \frac{2}{{\left( {2\sqrt {29} } \right)\left( {\sqrt {14} } \right)}} \cr
& \cos \theta = \frac{1}{{\sqrt {29} \sqrt {14} }} \cr
& {\text{solving for }}\theta \cr
& \theta = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt {29} \sqrt {14} }}} \right) \cr
& {\text{simplify by using a calculator}} \cr
& \theta \approx {87.2^ \circ } \cr} $$