Step 1

Rationalization. Geometrically speaking (and as taught in the book as a corollary) , two lines \(l_1\) and \(l_2 \) are parallel if and only if the cross products of their directional vectors a and b is the zero vector \(<0,0,0>=\vec{0}\). That is:

\(a \times b =\vec{0}\)

Two lines are intersecting if there exists a point (x,y,z) that is common in their domain . Lastly , two lines are skew if they are neither parallel or intersecting. We are given two sets of symmetric equations:

\(\frac{x}{1}=\frac{y-1}{-1}=\frac{z-2}{3}\)

\(\frac{x-2}{2}=\frac{y-3}{-2}=\frac{z}{7}\)

Two lines must be one of the three (parallel, intersecting, or skew)

Step 2

Acquiring the directional vectors. ‘he directional vectors of a parametric equation correspond to the denominators of each line. Thus, we get:

\(\frac{x}{1}=\frac{y-1}{-1}=\frac{z-2}{3} \Rightarrow v_1=<1,-1,3>\)

\(\frac{x-2}{2}=\frac{y-3}{-2}=\frac{z}{7} \Rightarrow v_2=<2,-2,7>\)

Step 3

Checking if the vectors are parallel. We can check if the vectors are parallel through the cross product or by ratio and proportion. The book suggests using the cross product method corollary, so let’s do just that:

\(v_1 \times v_2 = \begin{vmatrix}i & j &k \\1 & -1&-3\\2&-2&7 \end{vmatrix}\)

\(=\begin{vmatrix}-1 & -3 \\-2 & 7 \end{vmatrix}i-\begin{vmatrix}1 & -3 \\2 & 7 \end{vmatrix}j+\begin{vmatrix}1 & -1 \\2 & -2 \end{vmatrix}k\)

\(=[(-1)(7)-(-3)(-2)]i-[(1)(7)-(-3)(2)]j \)

\([(1)(-2)-(-1)(2)]k\)

\(=-13i-13j+0k \neq 0\)

Therefore, the lines are not parallel.

Step 4

Checking if the vectors are intersecting: rationalization. A common way of doing this is by equating x and y and solving for parameters s and t. . However, the given equations are still in symmetric form. Let's rewrite these symmetric equations into parametric form by equating to parametrs s and t ,respectively:

\(L_1: t=\frac{x}{1}=\frac{y-1}{-1}=\frac{z-2}{3} \Rightarrow \begin{cases}x=t\\y=-t+1\\z=3t+2\end{cases}\)

\(L_2: s=\frac{x-2}{2}=\frac{y-3}{-2}=\frac{z}{7} \Rightarrow \begin{cases}x=2s+2 \\ y=-2s+3 \\ z=7s \end{cases}\)

Substituting s and t to the parametric equations for z must yield an equality. We are checking the consistency of the linear system. Doing that, we get:

\(\begin{cases}x=t\\y=-t+1\\z=3t+2\end{cases}=\begin{cases}x=2s+2\\y=-2s+3\\z=7s\end{cases}\)

\(t=2s+2 \ \ \ \ (1)\)

\(-t+1=-2s+3 \ \ \ \ (2)\)

\(3t+2=7s \ \ \ \ \ \ (3)\)

Step 5

Checking if the vectors are intersecting: solving. Let’s solve for s and £ using equations (1) and (2). Doing that, we get:

\(\begin{cases}t&=2s+2&&t-2s&=2(4)\\&& \Rightarrow&&\\-t+1&=-2s+3&&-t+2s&=2(5)\end{cases} \)

Equating (4) and (5) by the elimination of t, we get:

\(\begin{cases}t-2s = 2 (4) \\ -t+2s=2(5) \end{cases} \Rightarrow t-t-2s+2s=2+2\)

There is a contradiction since \(0 \neq 4\). Therefore, this is an inconsistent system as there is no solution to s and t. Therefore, the lines are not intersecting.

Step 6

Conclusion.There are no values of s and t such that both equations are equal due to the inconsistency of the system. Therefore, the systems are also not intersecting. Therefore, the lines are skew.