Problem 34 - Advanced Engineering Mathematics

Question

In Problems, determine whether the given lines intersect. If so, find the point of intersection. $$ \begin{aligned} &x=3-t, y=2+t, z=8+2 t \\ &x=2+2 s, y=-2+3 s, z=-2+8 s \end{aligned} $$

Step-by-Step Solution

Answer

The lines intersect at the point \((4, 1, 6)\).

Step 1: Parametrization Review

The first line is given by the parametric equations \(x=3-t\), \(y=2+t\), and \(z=8+2t\). The second line is given by the parametric equations \(x=2+2s\), \(y=-2+3s\), and \(z=-2+8s\). Our goal is to find a common point \((x, y, z)\) that satisfies both sets of equations.

Step 2: Equating x-Coordinates

Set the x-coordinates equal to each other to solve for either \(t\) or \(s\), so we have \(3-t = 2+2s\). Rearrange this to get \(t = 1 - 2s\).

Step 3: Substituting t in y-Coordinates

Substitute \(t = 1 - 2s\) into the y-coordinates equation: \(y = 2+(1-2s) = 3-2s\). Set this equal to the y-coordinate of the second line: \(3-2s = -2+3s\).

Step 4: Solving y Equation

Solve \(3-2s = -2+3s\) for \(s\). Add \(2s\) to both sides: \(3 = -2+5s\). Add 2 to both sides: \(5 = 5s\). Finally, divide by 5: \(s = 1\).

Step 5: Determining t from s

Using \(s = 1\), substitute it back into \(t = 1 - 2s\) to find \(t\): \(t = 1 - 2(1) = -1\).

Step 6: Verifying with z-Coordinates

Verify that \(s = 1\) and \(t = -1\) satisfy the z-coordinate equations. For the first line: \(z = 8 + 2(-1) = 6\). For the second line: \(z = -2 + 8(1) = 6\). Both equal 6, confirming we have the correct values.

Step 7: Conclusion

The values \(t = -1\) and \(s = 1\) satisfy all the equations, indicating the lines intersect. Substitute \(t = -1\) into the first line's parametric equations to find the intersection point \((x, y, z) = (3-(-1), 2+(-1), 8+2(-1)) = (4, 1, 6)\).

Key Concepts

Parametric Equations

Parametric equations are an essential tool for describing lines in three-dimensional space. Unlike a simple slope-intercept form you see in two dimensions, parametric equations break down a point on the line into coordinates that depend on one or more parameters.
Consider the parametric equations for a line in three-dimensional geometry:

• For the first line: \(x = 3 - t\), \(y = 2 + t\), \(z = 8 + 2t\)
• For the second line: \(x = 2 + 2s\), \(y = -2 + 3s\), \(z = -2 + 8s\)

Here, the parameters \(t\) and \(s\) help us pinpoint any position along each respective line by varying their value.
These equations are particularly useful for finding intersections since you can equate them to form a system that, when solved, gives points common to both lines.
By manipulating the parameters, one can effectively trace every point along a line. Understanding how to apply parametric equations can greatly facilitate working with three-dimensional equations and their intersections.

System of Equations

The intersection of two lines can be effectively determined by forming a system of equations using their parametric equations. When investigating the intersection of lines, like in our original problem, we equate the parametric equations for each of the coordinates—\(x\), \(y\), and \(z\).
For example, by setting \(x = 3 - t = 2 + 2s\), we can solve for one parameter with respect to the other. This leads to our first simple equation: \(t = 1 - 2s\).
From here, you plug this expression of \(t\) into the equation for \(y\), forming the equation \(3 - 2s = -2 + 3s\), simplifying the problem algebraically to find \(s = 1\).
At this stage, solving this system of equations not only provides potential points of intersection but also confirms whether such a point exists. This is done by verifying the solution against the whole set of parametric equations, including the \(z\) coordinate equations. If all solutions are consistent, it confirms the lines intersect.

Three-Dimensional Geometry

Three-dimensional geometry involves exploring spaces that go beyond the typical length and width. A third coordinate—usually referred to as \(z\)—adds depth, raising the complexity but also the richness of geometric exploration.
In problems that involve lines in three dimensions, we work with equations such as \(x = 3 - t\), \(y = 2 + t\), and \(z = 8 + 2t\). These equations give a vector that can be visualized as traveling through space, bent and tilted by the parameters \(t\) and \(s\).

• Intersections in three dimensions are not as straightforward since lines can cross without actually intersecting unless they are on the same plane.
• Finding intersection points truly tests our understanding of space, because you must solve a series of equations simultaneously.

Understanding these aspects of three-dimensional geometry is critical for solving intersection problems. It helps you visualize and solve more complex problems accurately, knowing exactly where and how different geometric elements may interact.