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{\bf Question}

Given $\vec{OA} = (1,0,-1)$, $\vec{OB} = (1,2,3)$ and $\vec{OC} =
(0,1,2)$.  Find
\begin{description}
\item[(a)] the direction vector through $A$ and $B$;
\item[(b)] the vector equation of the line $AB$;
\item[(c)] the Cartesian equation of the line $AB$.
\end{description}

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{\bf Answer}

$$\vec{OA} = (1,0,-1), \hspace{.1in} \vec{OB} = (1,2,3),
\hspace{.1in} \vec{OC} = (0,1,2)$$
\begin{description}
\item[(a)]
$\vec{AB} = \vec{OB} - \vec{OA} = (1,2,3) - (1,0,-1) = (0,2,4) =
{\bf m}$
\item[(b)]
vector equation of the line AB is \begin{eqnarray*} {\bf r} & = &
(1,0,-1) + \lambda(0,2,4) \\ & = & (1, 2\lambda, -1+ 4 \lambda) \\
& = & {\bf r}_0 + \lambda {\bf m} \end{eqnarray*}
\item[(c)]
Cartesian equation is

$\begin{array}{rcl} x - x_0 & = & \lambda m_x \\ y - y_0 & = &
\lambda m_y \\ z - z_0 & = & \lambda m_z
\end{array}\ \  \Rightarrow\ \  \begin{array}{rcccl} x & = & 1 \\ y - y_0
& = & 2\lambda & = & y \\ z - z_0 & = & 4\lambda & = & z + 1
\end{array}$

Hence $(z + 1) = 2(2\lambda) = 2y$

So ${\underline {x = 1, z - 2y + 1 = 0}}$ are the equations of the
line.
\end{description}



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