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{\bf Question}

\begin{description}
\item[(a)] If ${\bf x \cdot x} = \alpha^2$ show that the length of
${\bf x}$ is fixed but its direction is arbitrary.  Hence solve
for ${\bf x}$ $$a{\bf x \cdot x} + 2{\bf b \cdot x} + c = 0$$
\item[(b)] Eliminate ${\bf y}$ from the equations
\begin{eqnarray*} p{\bf x - a \times y} & = & {\bf c} \\ {\bf b
\times x + y} & = & {\bf 0} \end{eqnarray*} Hence solve that this
pair of simultaneous equations for ${\bf x}$ and ${\bf y}$ when
${\bf a, b, c}$ are linearly independent, considering the 3 cases
$p \not= {\bf a \cdot b},$

$p = {\bf a \cdot b}, p = 0$.
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)] If $x \cdot x = \alpha ^2$ then $|x| = |\alpha|$ so the magnitude of $x$ is
fixed. \begin{eqnarray*} a\left(x \cdot x + \frac{2}{a} b \cdot x
+ \frac{c}{a} \right) & = & 0 \\ \left({\bf x} + \frac{b}{a}
\right) \cdot \left( {\bf x} + \frac{b}{a} \right) & = & \frac{b^2
- ac}{a^2} \end{eqnarray*} So $$x = \frac{1}{a}{\bf b} +
\sqrt{\frac{b^2 - ac}{a^2}} \hat{e}$$ where $\hat{e}$ is an
arbitrary unit vector provided $b^2 \geq ac$


\item[(b)]\begin{eqnarray*} p{\bf x - a \times y} & = & {\bf c} \\ {\bf b
\times x + y} & = & {\bf 0} \\ {\rm So} \hspace{.6in} p{\bf x + a
\times (b \times x) } & = & {\bf c} \\ p{\bf x} + (a \cdot x) {\bf
b} - (a \cdot b) {\bf x} & = & {\bf c} \\ {\rm or \ \ \ \ \ } (p -
a \cdot b){\bf x} + (a \cdot x) {\bf b} & = & {\bf c}
\hspace{.2in}(*) \end{eqnarray*}

This is the same equation as question 15.

If $p-{\bf a \cdot b} \not=0$ then $\ds {\bf x} = \frac{c - (a
\cdot x){\bf b}}{p - a \cdot b}$

Take the scalar product of (*) with ${\bf b}$ to find $\ds (a
\cdot x) = \frac{a \cdot c}{p} \hspace{.1in} p \not=0$

So $$x = \frac{{\bf c} - \frac{a \cdot c}{p} {\bf b}}{p - {\bf a
\cdot b}}$$

${}$

If $p-{\bf a \cdot b} =0$ then (*) gives $(a \cdot x){\bf b} =
{\bf c}$ and there are no solutions since ${\bf b}$ and ${\bf c}$
are map.

${}$

If $p=0$ then the scalar of (*) with ${\bf a}$ gives ${\bf a \cdot
c} = 0$ so there are solutions only if ${\bf a \cdot c} = 0$, then
we have $${\bf x} = \frac{c - (a \cdot x){\bf b}}{-a \cdot b}
\hspace{.2in} {\rm if \ } a \cdot b \not=0 \hspace{.2in}{bf x} =
-\frac{c}{a \cdot b} + \alpha {\bf b} \hspace{.1in}
\alpha\epsilon{\bf R}$$

${}$

If $p=0$ and $a \cdot b = 0$ then $(a \cdot x){\bf b = c}$ then
there are no solutions.

${}$

We then find ${\bf y}$ from ${\bf y = -b \times x}$

\end{description}

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