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{\bf Question}

Explain how the equation of a cylinder, of radius $a$ whose axis
in the direction of ${\bf \hat{n}}$ passes through$B$, can be
written in any of the forms
\begin{description}
\item[(i)] $\left| {\bf (r - b) \times \hat{n}}\right| = a.$
\item[(ii)] ${\bf (r - b) \times \hat{n}} = a{\bf \hat{e}}$

where ${\bf \hat {e}}$ is a unit vector orthogonal to ${\bf
\hat{n}}$.
\item[(iii)] ${\bf (r - b) } = a{\bf \hat{R} } + t{\bf \hat{n}}
\hspace{.2in} t\epsilon{\bf R}$

where ${\bf \hat{R} }$ is a unit vector perpendicular to ${\bf
\hat{n}}$  such that ${\bf \hat{R} }, -{\bf \hat{e}}, {\bf
\hat{n}}$ form a righthanded system.

\end{description}

\vspace{.25in}

{\bf Answer}

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${}$

${\bf r - b}$ is perpendicular to $n$ So $| r - b| =a$.

So $(r - b) \times n = a \hat{e}$

or $|(r - b) \times n| = a$

Also $|{\bf r - b} = a \hat{R} + t \hat{n}$

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