Determine whether the lines given below intersect, are parallel, or are skew.
Suppose the lines intersect. Then there is some value of the parameter \( t_{1} \) and some value of the parameter \(t_{2}\) for which \(\mathbf{r}_{1}= \mathbf{r}_{2}.\) Since \(~\mathbf{r}_{1}=(3+t_{1})\mathbf{i}+(3-2t_{1}) \mathbf{j}+(1+t_{1})\mathbf{k}\) and \(\mathbf{r}_{2}=(5+t_{2})\mathbf{i} +(1-4t_{2})\mathbf{j}+(1+3t_{2})\mathbf{k,}\) for \(\mathbf{r}_{1}\) to equal \( \mathbf{r}_{2},\) we have \[ \begin{equation*} (3+t_{1})\mathbf{i}+(3-2t_{1})\mathbf{j}+(1+t_{1})\mathbf{k}=(5+t_{2}) \mathbf{i}+(1-4t_{2})\mathbf{j}+(1+3t_{2})\mathbf{k}\qquad {\color{#0066A7}{\hbox{\(\mathbf{r}_{1}=\mathbf{r}_{2}\)}}} \end{equation*}
We equate the components of each vector to obtain \[ 3+t_{1}=5+t_{2} \qquad 3-2t_{1}=1-4t_{2} \qquad 1+t_{1}=1+3t_{2} \]
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The result is a system of three equations containing two variables. From the third equation, \(t_{1}=3t_{2}\). Now substitute \(t_{1}=3t_{2}\) into each of the first two equations. \[ \begin{equation*} \begin{array}{cllll} \begin{array}{rcl} 3+t_{1}&=&5+t_{2}\\ 3+3t_{2}&=&5+t_{2}\\ t_{2}&=&1 \end{array} & \ \ \ \ & \begin{array}{rcl} 3-2t_{1}&=&1-4t_{2}\\ 3-6t_{2}&=&1-4t_{2}\\ t_{2}&=&1 \end{array} & \ \ \ \ & {\color{#0066A7}{\hbox{\(t_{1}=3t_{2}\)}}} \end{array} \end{equation*}
Backsubstituting, we conclude \(t_{1}=3\) and \(t_{2}=1\). The point of intersection is found by substituting \(t_{1}=3\) in \(l_{1}\) or \(t_{2}=1\) in \(l_{2}\) . The result is \(\mathbf{r}_{1}=\mathbf{r}_{2}=6\mathbf{i}-3\mathbf{j}+4 \mathbf{k}\). The point of intersection is \((6,-3,4)\).
(b) The line \(l_{1}\) is in the direction of the vector \(\mathbf{i}- \mathbf{j}+\mathbf{k,}\) and the line \(l_{2}\) is in the direction of the vector \(\mathbf{i}-\mathbf{j}+2\mathbf{k}\). These vectors are not parallel, so the lines \(l_{1}\) and \(l_{2}\) either intersect or are skew. As before, suppose they intersect. Then \(\mathbf{r}_{1}=\mathbf{r}_{2}\mathbf{\ }\)so that \[ \begin{eqnarray*} &&( 3+t_{1}) \mathbf{i}+\left( 2-t_{1}\right)\! \mathbf{j}+( 1+t_{1}) \mathbf{k}=( 5+t_{2})\mathbf{i}+( 6-t_{2})\mathbf{j}+( 1+2t_{2}) \mathbf{k}\\[4pt] &&3+t_{1}=5+t_{2} \qquad 2-t_{1}=6-t_{2} \qquad 1+t_{1}=1+2t_{2} \qquad {\color{#0066A7}{\hbox{Equate components.}}} \end{eqnarray*}
Now substitute \(t_{1}=2t_{2}\) (from the third equation) into each of the first two equations. \[ \begin{equation*} \begin{array}{cllll} \begin{array}{rcl} 3+t_{1}&=&5+t_{2} \\[6pt] 3+2t_{2}&=&5+t_{2} \\[6pt] t_{2}&=&2 \end{array} & \ \ \ \ & \begin{array}{rcl} 2-t_{1}&=&6-t_{2} \\[6pt] 2-2t_{2}&=&6-t_{2} \\[6pt] t_{2}&=&-4 \end{array} & \ \ \ \ & {\color{#0066A7}{\hbox{\(t_{1}=2t_{2}\)}}} \end{array} \end{equation*} \]
The system of equations has no solution, so the assumption that \(\mathbf{r} _{1}=\mathbf{r}_{2}\mathbf{,}\) for some \(t_{1}\) and some \(t_{2}\) is false. The lines are neither parallel nor intersecting, so they are skew.