Question

A pair of lines in \(\mathbb{R}^{3}\) are said to be skew if they are neither parallel nor intersecting. Determine whether the following pairs of lines are parallel, intersecting, or skew. If the lines intersect. determine the point(s) of intersection. $$\begin{aligned} &\mathbf{r}(t)=\langle 1+2 t, 7-3 t, 6+t\rangle;\\\ &\mathbf{R}(s)=\langle-9+6 s, 22-9 s, 1+3 s\rangle \end{aligned}$$

Short answer

Answer: The lines intersect at the point \(\mathbf{P} = \langle -5, 16, 3 \rangle\).

Step-by-step solution

Check for parallelism

To check if the given lines are parallel, we need to see if their direction vectors are proportional. The direction vector of \(\mathbf{r}(t)\) is the derivative of the given equation with respect to \(t\). Similarly, the direction vector of \(\mathbf{R}(s)\) is the derivative of the given equation with respect to \(s\). The direction vector of \(\mathbf{r}(t)\) is: $$\mathbf{v}_{r} = \frac{d}{dt}\mathbf{r}(t) = \langle 2, -3, 1 \rangle$$ The direction vector of \(\mathbf{R}(s)\) is: $$\mathbf{v}_{R} = \frac{d}{ds}\mathbf{R}(s) = \langle 6, -9, 3 \rangle$$ Since \(\mathbf{v}_{R}\) is three times \(\mathbf{v}_{r}\), the lines are parallel or intersecting.

Check for intersection or parallelism

We need to find if there is a point shared by both lines. We can do this by setting the line equations equal to each other and solving for the parameters \(t\) and \(s\). $$ \mathbf{r}(t) = \mathbf{R}(s)$$ Which, when expanded, yields a system of three equations: $$ \begin{cases} 1+2t=-9+6s \\ 7-3t=22-9s \\ 6+t=1+3s \end{cases} $$ Solving this system of equations, we find that \(t=-3\) and \(s=1\), which yields the same point for both \(\mathbf{r}(t)\) and \(\mathbf{R}(s)\): $$\mathbf{r}(-3) = \langle 1 + 2(-3),7-3(-3),6-3\rangle=\langle -5, 16, 3\rangle$$ $$\mathbf{R}(1) = \langle -9 + 6(1), 22 - 9(1), 1 + 3(1)\rangle= \langle -5, 16, 3 \rangle$$ Since there is a common point, the lines intersect at the point: $$\mathbf{P} = \langle -5, 16, 3 \rangle$$

Key concepts

Intersection of Lines

To determine if two lines intersect, you set the equations of the two lines equal and solve for the parameters. These lines are defined in vector form. For example, line \( \mathbf{r}(t) \) is defined with parameter \( t \) and line \( \mathbf{R}(s) \) with parameter \( s \). To find intersection:

• Write the equations as a system.
• Equate the respective components.
• Solve the system to see if there are values of \( t \) and \( s \) that plug into both line equations to yield the same point.

When comparing like components, if you achieve a consistent set of \( t \) and \( s \), the lines intersect at that point. In our exercise, solving the equations found \( t = -3 \) and \( s = 1 \). Substituting these back into the line equations confirms the lines intersect at point \( \mathbf{P} = \langle -5, 16, 3 \rangle \).

Direction Vectors

A direction vector of a line indicates its direction in space. For a parametric line equation like \( \mathbf{r}(t) = \langle 1+2t, 7-3t, 6+t \rangle \), the direction vector \( \mathbf{v}_r \) can be found by differentiating each component with respect to the parameter \( t \). Thus, \( \mathbf{v}_r = \langle 2, -3, 1 \rangle \).

• Direction vectors help identify if lines are parallel. Parallel lines possess proportional direction vectors.
• It's important to check vectors for proportionality to assess parallelism before further examining intersection possibilities.
• If direction vectors are scalar multiples of each other, lines are parallel or might intersect if sharing a point.

In our exercise, the vector \( \mathbf{v}_R = \langle 6, -9, 3 \rangle \) is three times \( \mathbf{v}_r \), suggesting the lines could be parallel or coinciding (if intersecting).

System of Equations

Systems of equations arise when you need to solve multiple equations simultaneously, often encountered when assessing intersection of lines. After setting the line equations equal component-wise:

• You form a system corresponding to each component equation.
• Solve to find values of parameters, revealing potential intersection points.
• Checking consistent solutions across all equations confirms genuine intersection instead of mere coincidence on a subset plane.

In the exercise, solving:\[\begin{cases}1 + 2t = -9 + 6s \7 - 3t = 22 - 9s \6 + t = 1 + 3s\end{cases}\]led to \( t = -3 \) and \( s = 1 \), demonstrating the system's solution as consistent. Thus, ensuring line \( \mathbf{r}(t) \) and \( \mathbf{R}(s) \) indeed intersect at point \( \mathbf{P} = \langle -5, 16, 3 \rangle \). Using this process confirms the problem's solution is thorough and complete.