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{\bf Question}

Let ${\bf a}$, ${\bf b}$ and ${\bf r}$ be vectors with ${\bf
a}\cdot {\bf b}\ne 0$, and let $t$ be a scalar. Show that the
equation: $${\bf a}\times {\bf r}={\bf a}+t{\bf b}$$ can be
satisfied only if $$t=-\frac{{\bf a}\cdot{\bf a}}{{\bf a}\cdot{\bf
b}}.$$ Put this value of $t$ into the equation and deduce that
${\bf r}$ must have the form:

$${\bf r}=\frac{{\bf a}\times {\bf b}}{{\bf a}\cdot {\bf b}} +
\left ( \frac{{\bf a}\cdot {\bf r}}{{\bf a}\cdot {\bf a}}
\right){\bf a}.$$



{\bf Answer}

${\bf a} \times {\bf r} = {\bf a} + t{\bf b}$ \hspace{.25in}(*)

Take the dot products of both sides of (*) with ${\bf a}$:

$({\bf a} \times {\bf r}) \cdot {\bf a} = ({\bf a} + t{\bf b})
\cdot {\bf a}={\bf a}\cdot{\bf a}+t{\bf b}\cdot{\bf a}$

and since $({\bf a} \times {\bf r}) \cdot {\bf a} = 0$ we have\ $
0 = {\bf a}\cdot{\bf a}+t{\bf b}\cdot{\bf a}$ or\ $ {\bf a} \cdot
{\bf b} t = - {\bf a} \cdot {\bf a}$

Dividing by the scalar $ {\bf a} \cdot {\bf b}$ gives
$\displaystyle t = \frac { - {\bf a} \cdot {\bf a}}{{\bf a} \cdot
{\bf b} }$

Substitute this value of $t$ into (*) to obtain:

$$ {\bf a} \times {\bf r} = {\bf a} + \left( \frac { - {\bf a}
\cdot {\bf a}}{{\bf a} \cdot {\bf b} } \right) {\bf b} {\rm
\hspace{.25in} (**)}$$

Take the cross product of both sides of (**) with ${\bf a}$:

\begin{eqnarray*}
{\bf a}\times ({\bf a} \times {\bf r}) & = & {\bf a} \times \left(
{\bf a} - \left( \frac {{\bf a} \cdot {\bf a}}{{\bf a} \cdot {\bf
b} } \right) {\bf b} \right)\\ & = & {\bf a} \times {\bf a}
-\left( \frac {{\bf a} \cdot {\bf a}}{{\bf a} \cdot {\bf
b}}\right) ({\bf a} \times {\bf b}){\rm \ \ \ (Distributive\
Property)}\\ & = & -\left( \frac {{\bf a} \cdot {\bf a}}{{\bf a}
\cdot {\bf b}}\right) ({\bf a} \times {\bf b}){\rm \ \ \ (since\
{\bf a} \times {\bf a} = 0)}
\end{eqnarray*}

From question 3 we have:
\begin{eqnarray*}
{\bf a} \times ({\bf a} \times {\bf r}) & = & ({\bf a} \cdot {\bf
r}) {\bf a} - ({\bf a} \cdot {\bf a}){\bf r}\\ {\rm and\ so\ \ \ }
({\bf a} \cdot {\bf r}){\bf a} - ({\bf a}\cdot{\bf a}){\bf r} & =
& -\left( \frac{{\bf a} \cdot {\bf a}}{{\bf a} \cdot {\bf b}}
\right)({\bf a} \times {\bf b}).
\\ {\rm Rearranging\ gives\ \ \ } {\bf r} & = & \frac{{\bf a}
\times {\bf b}}{{\bf a} \cdot {\bf b}} + \left( \frac{{\bf a}
\cdot {\bf r}}{{\bf a} \cdot {\bf a}} \right){\bf a}\\ ({\rm
assume\ {\bf a}\ is\ a\ non-zero\ vector,\ so\ \ } {\bf a} \cdot
{\bf a} & = & |{\bf a}|^2 \not= 0)
\end{eqnarray*}


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