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\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
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\begin{document}

{\bf Question}

Find the vector equation of the line through the point with
position vector ${\bf{a}}=2{\bf{i}}-{\bf{j}}-3{\bf{k}}$ which is
parallel to the vector ${\bf{b}}={\bf{i}}+{\bf{j}}+{\bf{k}}$.
Determine the points corresponding to $\lambda=3,0,2$ in resulting
equation. Write down the parametric and cartesian equations of the
line.
\medskip

{\bf Answer}

${}$

\setlength{\unitlength}{.5in}
\begin{picture}(5,3)
\put(0,0){\vector(1,1){2}}

\put(0,0){\vector(1,0){5}}

\put(5,0){\line(-3,2){4}}

\put(0,0){\circle*{.125}}

\put(1,1){\makebox(0,0)[r]{${\bf{a}}$}}

\put(3.5,-.25){\makebox(0,0)[b]{${\bf{r}}$}}

\put(4,1){\makebox(0,0)[b]{${\bf{b}}$}}

\put(3,2){\vector(3,-2){2}}
\end{picture} \vspace{.2in}

Vector equation:
${\bf{r}}={\bf{a}}+\lambda{\bf{b}}=2{\bf{i}}-{\bf{j}}-3{\bf{k}}+
\lambda({\bf{i}}+{\bf{j}}+{\bf{k}})$

\un{$\lambda=-3$}:
${\bf{r}}=2{\bf{i}}-{\bf{j}}-3{\bf{k}}-3({\bf{i}}+{\bf{j}}+{\bf{k}})
=-{\bf{i}}-4{\bf{j}}-6{\bf{k}}$

\un{$\lambda=0$}:
${\bf{r}}=2{\bf{i}}-{\bf{j}}-3{\bf{k}}+0({\bf{i}}+{\bf{j}}+{\bf{k}})
=2{\bf{i}}-{\bf{j}}-3{\bf{k}}={\bf{a}}$

\un{$\lambda=2$}:
${\bf{r}}=2{\bf{i}}-{\bf{j}}-3{\bf{k}}+2({\bf{i}}+{\bf{j}}+{\bf{k}})
=4{\bf{i}}+{\bf{j}}-{\bf{k}}$

Parametric equation:

${\bf{r}}=2{\bf{i}}-{\bf{j}}-3{\bf{k}}+\lambda({\bf{i}}+{\bf{j}}+{\bf{k}})$

$=(2+\lambda){\bf{i}}+(\lambda-1){\bf{j}}+(\lambda-3){\bf{k}}$

General point is $(x,y,z)$, so

$$\left\{\begin{array}{l} x=2+\lambda\\y=\lambda-1\\z=\lambda-3
\end{array} \right\}$$ Parametric equation

Cartesian equation: $\lambda$ must be the same for each of the
parametric equations. Thus eliminating $\lambda$,

$x-2=\lambda,\ y+1=\lambda,\ z+3=\lambda \Rightarrow
\un{x-2=y+1=z+3}$

\end{document}