Question

Consider the lines $$\begin{aligned}\mathbf{r}(t) &=\langle 2+2 t, 8+t, 10+3 t\rangle \text { and } \\\\\mathbf{R}(s) &=\langle 6+s, 10-2 s, 16-s\rangle\end{aligned}$$. a. Determine whether the lines intersect (have a common point) and if so, find the coordinates of that point. b. If \(\mathbf{r}\) and \(\mathbf{R}\) describe the paths of two particles, do the particles collide? Assume that \(t \geq 0\) and \(s \geq 0\) measure time in seconds, and that motion starts at \(s=t=0\).

Short answer

**Answer:** The lines intersect at the point \(\left(\frac{18}{5}, \frac{42}{5}, \frac{56}{5}\right)\). However, the particles do not collide since the condition \(t, s \geq 0\) is not met for \(s\).

Step-by-step solution

Write down the position vectors of the lines

We are given the position vectors of the lines as: $$\begin{aligned}\mathbf{r}(t) &=\langle 2+2 t, 8+t, 10+3 t\rangle \\\mathbf{R}(s) &=\langle 6+s, 10-2 s,16-s\rangle\end{aligned}$$

Equate the coordinates of the lines

For the lines to intersect, their coordinates must be equal for some values of \(t\) and \(s\): $$\begin{cases} 2+2t = 6+s \\ 8+t = 10-2s \\ 10+3t = 16-s \end{cases}$$

Solve the system of equations

To solve the system of equations, we first solve the first equation for \(s\) and then substitute to the other equations until we find a unique solution for \(t\) and \(s\). First equation: \(s = 2t - 4\) Second equation: \(8+t = 10-2(2t-4) \Rightarrow t = 2(2t-4)\) Third equation: \(10+3t = 16-(2t-4)\Rightarrow t = \frac{2}{5}\) Now, we can substitute the value of \(t\) back into the first equation to find the value of \(s\): \(s = 2 \cdot \frac{2}{5} - 4 = -\frac{28}{5}\)

Determine if the lines intersect

Since we have found values of \(t\) and \(s\) that make the coordinates equal, the lines intersect. We can find the intersection point by plugging the value of \(t\) into the position vector \(\mathbf{r}(t)\): $$\mathbf{r}\left(\frac{2}{5}\right) = \left\langle 2 + 2 \cdot \frac{2}{5}, 8 + \frac{2}{5}, 10 + 3 \cdot \frac{2}{5} \right\rangle = \left\langle \frac{18}{5}, \frac{42}{5}, \frac{56}{5} \right\rangle$$ Thus, the lines intersect at point \(\left(\frac{18}{5}, \frac{42}{5}, \frac{56}{5}\right)\).

Determine if the particles collide

We now have to determine if the particles collide. For the particles to collide, the values of \(t\) and \(s\) we found must be non-negative, since we are given that \(t, s \geq 0\). We have: $$t = \frac{2}{5} \geq 0$$ However, we also have: $$s = -\frac{28}{5} < 0$$ Since \(s < 0\), the particles will not collide.

Key concepts

Vector Equations

A vector equation represents a line or curve in space, expressing the position of any point on the line as a vector function of a single parameter. In the given exercise, we work with vector equations to identify intersecting lines. When we say \( \mathbf{r}(t) = \langle 2+2t, 8+t, 10+3t \rangle \) and \( \mathbf{R}(s) = \langle 6+s, 10-2s, 16-s \rangle \), we're representing two lines in three dimensions using position vectors.

For example, the parameter \( t \) in the equation of \( \mathbf{r}(t) \) can represent time, meaning as time progresses, the point on the line moves, indicating motion. By analyzing vector equations, we determine the behavior of the lines, including points of intersection which can be very useful in fields like physics and engineering where space and motion are key elements of study.

Systems of Equations

A system of equations consists of multiple equations that are solved together. Addressing these systems is crucial to finding the intersection points of lines in three dimensions. In our exercise, we're presented with a three-equation system derived from equating the corresponding components of our two position vectors. Solving for \( t \) and \( s \) allows us to find specific points in space where the lines might intersect.

To find a solution, we can employ various techniques such as substitution, elimination, or use matrix methods like row reduction. It's essential to grasp these strategies since they form the foundation for solving more complex real-world problems involving intersections and overlaps in space.

Particle Motion in Space

Examining particle motion in space involves understanding how objects travel through a three-dimensional environment over time. In our context, the vector equations \( \mathbf{r}(t) \) and \( \mathbf{R}(s) \) illustrate the paths of two particles. To determine if they collide, we need to find a common point where their paths intersect at the same time.

We equate the vectors for potential points of intersection and scrutinize the corresponding parameter values to check for a collision. In practical terms, this can represent anything from atoms to astronomical bodies, making it relevant for studies in both microscopic and cosmic scales.

Position Vectors

Position vectors describe the location of a point in space relative to an origin. The expressions \( \mathbf{r}(t) \) and \( \mathbf{R}(s) \), from our exercise, are position vectors. They provide a directional and magnitude component conveying where the point (or particle) is at any instant when parameterized by \( t \) or \( s \).

In different applications such as robotics, aviation, or game development, understanding position vectors is essential for navigating and controlling movement within a three-dimensional environment.