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{\bf Question}

When $k{\bf x} + {(\bf x \cdot a)b} = {\bf c} \, \, (b \not= 0)$
explain why ${\bf x}$ is of the form $\frac{1}{k}{\bf c} +
\alpha{\bf b} , \, \, k \not=0 \, \, \alpha\epsilon{\bf R}$. Hence
solve  $k{\bf x} + {(\bf x \cdot a)b} = {\bf c}$
\begin{description}
\item[(i)]  $k + a \cdot b \not= 0$
\item[(ii)] $k + a \cdot b = 0$ and ${\bf a \cdot c} = 0$ or  ${\bf a \cdot c} \not=
0$.
\end{description}

What can you say about possible solutions if $k=0$?

\vspace{.25in}

{\bf Answer}

If $k{\bf x + (x \cdot a) b = c}$

Then ${\bf x} = \frac{1}{k} {\bf c} - \frac{1}{k}{\bf (x \cdot
a)b} = \frac{1}{k}{\bf c + \alpha b} \hspace{.2in} k \not=0 \, \,
\, \alpha \epsilon {\bf R}$

So substituting for ${\bf x}$ in the original equation gives

$k{\bf \frac{1}{k}{\bf c + \alpha b} + (\frac{1}{k}{\bf c + \alpha
b} \cdot a) b = c}$

$(k\alpha + \frac{1}{k}{\bf c \cdot a + \alpha b \cdot a) b = 0}$

So $\alpha(k + {\bf b \cdot a} = -\frac{1}{k} (c \cdot a)
\hspace{.2in} {\bf b \not=0}$

\begin{description}
\item[(i)] So $\ds \alpha = -\frac{1}{k} \frac{{\bf c \cdot a}}{k +
{\bf b \cdot a}} \hspace{.2in} {\rm if\ } k +{\bf b \cdot a} \not=
0$

Thus in this case we have $$ x = \frac{1}{k} {\bf c} - \frac{1}{k}
\frac{{\bf c \cdot a}}{k + {\bf b \cdot a}}{\bf b}$$

\item[(ii)] If $k + {\bf b \cdot a} = 0$ and ${\bf c \cdot a} = 0$
then any $\alpha$ will give a solution.

If $k + {\bf b \cdot a} = 0$ and ${\bf c \cdot a} = 0$ then are no
solutions.

If $k=0$ then the equation reduced to $(x \cdot a) {\bf b} = {\bf
c}$

If ${\bf b = c} = 0$ all ${\bf x}$ are solutions.

If only ${\bf b} =0$ then there are no solution.

If ${\bf c} = 0, \,  {\bf b} \not=0$ then ${\bf x \cdot a} =0$ so
${\bf x}$ and ${\bf a}$ are perpendicular.

If ${\bf c} \not= 0, \,  {\bf b} \not=0$ there are no solutions so
${\bf b}$ and ${\bf c}$ are not parallel.

If ${\bf b}$ and ${\bf c}$ are parallel then $\ds |(x \cdot a)| =
\frac{|c|}{|b|}$ i.e. $( x \cdot a) = \pm \frac{|c|}{|b|}$ as
${\bf b}$ and ${\bf c}$ have the same or opposite direction.  Thus
${\bf x} = \pm \frac{|c|}{|b||a|^2}{\bf a} + {\bf b}$ where ${\bf
b}$ and ${\bf a}$ are parallel.


\end{description}
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