Determine whether the lines given below intersect, are parallel, or are skew.

Solution (a) The line $l_{1}$ is in the direction of the vector $\mathbf{i}-2\mathbf{j}+\mathbf{k}$ and the line $l_{2}$ is in the direction of the vector $\mathbf{i}-4\mathbf{j}+3\mathbf{k}$. Since these vectors are not parallel (do you know why?), $l_{1}$ and $l_{2}$ either intersect 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 $$3+t_{1}=5+t_{2} \qquad 3-2t_{1}=1-4t_{2} \qquad 1+t_{1}=1+3t_{2}$$

737

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.