\documentclass[a4paper,12pt]{article}
\begin{document}


{\bf Question}

Find equations for the following lines in vector and scalar
parametric forms, and in standard form:
\begin{description}
\item[(a)]
the line that passes through the point $(0,1,0)$ and is parallel
to $\bf v = \bf i + \bf j - \bf k$;
\item[(b)]
the line that passes through the points $(1,2,2)$ and $(3,-1,3)$.

\end{description}


{\bf Answer}

\begin{description}
\item[(a)]
Equation of line in vector parametric form:

$$\left( \begin{array}{r} x\\ y\\ t
\end{array}  \right)=\left( \begin{array}{r}
     0\\ 1\\ 0
\end{array}  \right)+t\left( \begin{array}{r}
     1\\ 1\\ -1
\end{array}  \right)$$
\setlength{\unitlength}{.5in}

\begin{center}
\begin{picture}(3,2)
\put(1,0){\line(1,1){2}} \put(1.5,0.5){$\bullet$}
\put(0.2,0.5){$(0,1,0)$} \put(2,0.5){\vector(1,1){1}}
\put(3,1){$\left( \begin{array}{r} 1\\ 1\\ -1
\end{array}  \right)$}
\end{picture}
\end{center}

so that $x=t,\ y=1+t,\ z=-t$ and equation of line in scalar
parametric form is: $(x,y,z)=(t,1+t,-t)$.

Now,

$$ \left. \begin{array} {lll} {x = t} & {\Rightarrow} & {t = x}\\
                      {y = 1+t} & {\Rightarrow} & {t = y-1} \\
                      {z = -t} & {\Rightarrow} & {t = -z}
\end{array} \right \}.$$
So the equation in standard form is: $x=y-1=-z$

\item[(b)]
The line passes through the point $(1,2,2)$ and is parallel to the
vector: $$\left( \begin{array}{r}
     3\\
     -1\\
     3
\end{array}  \right)-\left( \begin{array}{r} 1\\ 2\\ 2
\end{array}  \right)=\left( \begin{array}{r} 2\\ -3\\ 1
\end{array}  \right)$$

\setlength{\unitlength}{.5in}

\begin{center}
\begin{picture}(4,2)
\put(1,0){\line(2,1){3}} \put(1.5,0.23){$\bullet$}
\put(2.5,.73){$\bullet$} \put(0.2,0.3){$(0,1,0)$}
\put(1.1,0.8){$(3,-1,3)$} \put(3,0.5){\vector(2,1){1}}
\put(4,0.5){$\left(
\begin{array}{r} 2\\ -3\\ 1
\end{array}  \right)$}
\end{picture}
\end{center}

Equation in vector parametric form: $$\left(
\begin{array}{r} x\\ y\\ t
\end{array}  \right)=\left( \begin{array}{r} 1\\ 2\\ 2
\end{array}  \right)+t\left( \begin{array}{r} 2\\ -3\\ 1
\end{array}  \right)$$
so that $x=1+2t,\ y=2-3t,\ z=2+t$ and equation in scalar
parametric form: $(x,y,z)=(1+2t,2-3t,2+t)$

$$ \left. \begin{array} {lll} {x = 1+2t} & {\Rightarrow} &
{\displaystyle t = \frac{x-1}{2}}\\
                      {y = 2-3t} & {\Rightarrow} & {\displaystyle t = \frac{2-y}{3}} \\
                      {z = 2+t} & {\Rightarrow} & {t = z-2}
\end{array} \right \}$$
So the equation in standard form:
$$\frac{x-1}{2}=\frac{2-y}{3}=z-2$$
\end{description}


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