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{\bf Question}

Given ${\bf u} = (4,2,0),  \hspace{.1in} {\bf v} =(-3,1,1),
\hspace{.1in} {\bf w} = (5,1,5), \hspace{.1in} {\bf s} = (1,2,1)$
find; (a) the angle between ${\bf u}$ and ${\bf w}$; (b) The value
of $\mu$ for which for ${\bf u} + \mu {\bf v}$ is perpendicular to
${\bf w}$


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{\bf Answer}

$${\bf u} = (4,2,0) \, {\bf v} =(-3,1,1) \, {\bf w} = (5,1,5) \,
{\bf s} = (1,2,1) $$
\begin{description}
\item[(a)]
${\bf u} \cdot {\bf w} = |{\bf u}| |{\bf w}| \cos \theta$ with
$\theta$ the angle between ${\bf u}$ and ${\bf w}$

${\bf u} \cdot {\bf w} = (4,2,0) \cdot (5,1,1) = 20 + 2 + 0 = 22$

$|{\bf u}| = \sqrt {4^2 + 2^2 + 0^2} = \sqrt{20}$

$|{\bf w}| = \sqrt {5^2 + 1^2 + 1^2} = \sqrt{27}$

Hence $\cos \theta = \frac{22}{\sqrt{540}} \approx 0.946
\Rightarrow \theta \approx 18.79 ^\circ$
\item[(b)]
$({\bf u} + \mu {\bf v})\cdot {\bf w} = 0$ for  ${\bf u} + \mu
{\bf v}$ perpendicular to ${\bf w}$

Hence \begin{eqnarray*} {\bf u} \cdot {\bf w} + \mu {\bf v} \cdot
{\bf w} & = & 0 \\ \mu & = & \frac{-{\bf u} \cdot {\bf w}}{{\bf v}
\cdot {\bf w}} \\ & = & \frac{-22}{(-3 \times 5 + 1 \times 1 + 1
\times 1)} \\ & = & \frac{22}{13} \end{eqnarray*}
\end{description}



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