Question

Find the points (if they exist) at which the following planes and curves intersect. $$z=16 ; \mathbf{r}(t)=\langle t, 2 t, 4+3 t\rangle, \text { for }-\infty < t < \infty$$

Short answer

Question: Find the point of intersection between the plane given by the equation \(z=16\) and the curve given by the vector-valued function \(\mathbf{r}(t)=\langle t, 2t, 4+3t \rangle\). Answer: The point of intersection between the plane and the curve is (4, 8, 16).

Step-by-step solution

Identify the equations for the plane and the curve

We are given the equation of the plane as \(z=16\) and the curve as \(\mathbf{r}(t)=\langle t, 2t, 4+3t \rangle\).

Substitute the curve into the plane equation

To find the points of intersection, we need to substitute the \(z\)-coordinate of the curve into the plane equation. From the curve equation, we have \(z = 4+3t\). Now, substitute this into the plane equation: $$16 = 4+3t$$

Solve for the parameter t

Now, we need to solve the equation for \(t\): $$16 - 4 = 3t$$ $$12 = 3t$$ $$t = \frac{12}{3} = 4$$

Find the point of intersection

Now that we have the value of \(t\), we can find the point of intersection. Plug the value of \(t=4\) into the curve equation: $$\mathbf{r}(4) = \langle 4, 2 \times 4, 4+3 \times 4 \rangle = \langle 4, 8, 16\rangle$$ So, the point of intersection between the plane and the curve is \((4,8,16)\).

Key concepts

Parametric Equations

In mathematics, parametric equations provide a way to express geometric objects using parameters. Rather than writing the coordinates solely in terms of variables like \(x, y,\) and \(z\), we use a parameter (often denoted as \(t\)). This allows us to describe more complex paths, such as curves, without being restricted to the Cartesian coordinate system.

The given curve in our exercise is represented by the vector function \( \mathbf{r}(t) = \langle t, 2t, 4+3t \rangle \). This means:

• \(x\) is expressed as \(t\)
• \(y\) is expressed as \(2t\)
• \(z\) is expressed as \(4 + 3t\)

This representation is powerful because it reveals how each coordinate changes as the parameter \(t\) varies. When solving problems like finding the intersection with a plane, this form allows us to substitute and find exact solutions for these points efficiently.

Coordinate Geometry

Coordinate geometry links algebra with geometry through the use of coordinates. It provides a way to calculate distances, angles, and other spatial relationships through algebraic formulas. In our exercise, we saw this come into play when we combined the parametric curve with the plane equation \(z = 16\).

This linking is essential for solving problems involving intersections. By substituting the expressions from the vector function into the plane's equation, we treat each coordinate as a linear equation. Solving \(16 = 4 + 3t\), we find \(t = 4\). This approach allows us to find exact points on the curve that meet certain geometric criteria outlined by the plane equation.

Coordinate geometry helps us translate abstract geometrical ideas into tangible solutions, providing both clarity and precision.

Vector Functions

Vector functions extend regular functions by using vectors, which include direction and magnitude. They're useful in physics and engineering to model anything that has both magnitude and direction, such as velocity or force.

In our scenario, the vector function \( \mathbf{r}(t) = \langle t, 2t, 4+3t \rangle \) defines the path of a curve in three-dimensional space, where each component depends on the parameter \(t\).

The advantage of using vector functions is that they allow an intuitive understanding of movement along paths. By substituting values like \(t = 4\), we find

• The specific point on the curve: \(\langle 4, 8, 16 \rangle\)
• Both the direction and rate at which a point moves along the curve

This method simplifies tracking and predicting behaviors of complex systems, providing a versatile tool in both theoretical and applied mathematics.