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{\bf Question}

\begin{description}
\item[(a)] For ${\bf c} = 5{\bf a} - {\bf b}$ and ${\bf d} = 3{\bf
a} + 2{\bf b}$ find ${\bf c} \cdot {\bf d}$ when
\begin{description}
\item[(i)] ${\bf a}$ and ${\bf b}$ are unit vectors at an angle
$\frac{\pi}{4}$
\item[(ii)]  ${\bf a}$ and ${\bf b}$ are perpendicular  with $|{\bf
b}| = 2|{\bf a}| = 2.$
\end{description}
\item[(b)] Evaluate $|{\bf c} \cdot {\bf d}|$ when ${\bf c} = {\bf
i} + 2{\bf j} + 3{\bf k}$ and ${\bf d} = 2{\bf i} - {\bf j} + {\bf
k}$

\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]
\begin{description}
\item[(i)] \begin{eqnarray*} {\bf c \cdot d} & = & (5{\bf a - b})
\cdot (3{\bf a} +2{\bf b} \\ & = & 15 {\bf a \cdot a} + 10{\bf a
\cdot b} - 3{\bf a \cdot b} -2{\bf b \cdot b} \\ & = & 13 +
\frac{7}{\sqrt2} \end{eqnarray*}
\item[(ii)]  \begin{eqnarray*} {\bf c \cdot d} & = & 15 {\bf a \cdot a} + 7{\bf a
\cdot b} -2{\bf b \cdot b} \\ & = & 15|{\bf a}|^2  - 2|{\bf b}|^2
\\ & = & 15 - 8 \\ & = & 7 \end{eqnarray*}
\end{description}
\item[(b)] ${\bf c} - (1,2,3) \hspace{.3in} {\bf d} = (2,-1,-1)$

${\bf c \cdot d} = 2-2-3 = -3$ so $|{\bf c \cdot d} | = 3$
\end{description}

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