Problem 57 - University Calculus: Early Transcendentals

Question

find the point in which the line meets the plane. $$x=1-t, \quad y=3 t, \quad z=1+t ; \quad 2 x-y+3 z=6$$

Step-by-Step Solution

Answer

The line meets the plane at the point \(\left(\frac{3}{2}, -\frac{3}{2}, \frac{1}{2}\right)\).

Step 1: Substitute the line parameters into the plane equation

To find the intersection, substitute the parameterized line equations into the plane equation. The line is given as \(x = 1 - t\), \(y = 3t\), \(z = 1 + t\). Substitute these into the plane equation \(2x - y + 3z = 6\): \[2(1-t) - 3t + 3(1+t) = 6.\]

Step 2: Simplify the equation

Simplify the equation obtained from substitution: \[2 - 2t - 3t + 3 + 3t = 6.\]Combine like terms:\[5 - 2t = 6.\]

Step 3: Solve for the parameter t

Now, solve the simplified equation for \(t\):\[5 - 2t = 6\]Subtract 5 from both sides:\[-2t = 1\]Divide by -2:\[t = -\frac{1}{2}.\]

Step 4: Find the intersection point

Using the value of \(t\), calculate the \(x\), \(y\), and \(z\) coordinates:For \(x\):\[x = 1 - \left(-\frac{1}{2}\right) = 1 + \frac{1}{2} = \frac{3}{2}.\]For \(y\):\[y = 3\left(-\frac{1}{2}\right) = -\frac{3}{2}.\]For \(z\):\[z = 1 + \left(-\frac{1}{2}\right) = \frac{1}{2}.\]So, the intersection point is \(\left(\frac{3}{2}, -\frac{3}{2}, \frac{1}{2}\right)\).

Key Concepts

Parameterized Line Equations

In the world of coordinate geometry, a parameterized line equation offers a versatile way to describe a line. Unlike the standard line equation form, parameterized equations express each coordinate as a function of a parameter, usually denoted as \( t \). This approach is particularly helpful in three-dimensional geometry, where lines are not confined to a plane.
For example, given the parameterized equations \( x = 1 - t \), \( y = 3t \), and \( z = 1 + t \), here each variable \( x, y, \text{ and } z \) is described as a function of \( t \).
Such a description allows us to find specific points on the line by simply assigning a value to \( t \).

• At \( t = 0 \), we find the initial point, \( (1, 0, 1) \).
• At any other \( t \), this set will yield a different point along the same line.

The flexibility of parameterized line equations is one of many reasons they are so useful in both pure and applied mathematics.

Equation of a Plane

Understanding the equation of a plane is crucial when dealing with intersections between lines and planes. A plane in three-dimensional space can be described using a linear equation of the form \( ax + by + cz = d \).
This represents a flat surface extending infinitely in two dimensions.
In our example, the plane is given by the equation \( 2x - y + 3z = 6 \). Here, the coefficients \( 2, -1, \text{ and } 3 \) can be seen as representing the orientation and tilt of the plane within the 3D space.

• The value \( d = 6 \) determines the plane’s position relative to the origin, essentially "moving" the plane outward.

Finding where a line intersects a plane helps us understand spatial relationships and can have practical applications in fields like computer graphics and engineering.

Solving for Parameter t

To discover where a parameterized line intersects a plane, we must solve for the parameter \( t \). This process involves substituting each parameterized coordinate equation into the plane's equation.
Let's go through the steps using our example:
1. Substitute: Take \( x = 1 - t \), \( y = 3t \), and \( z = 1 + t \) and substitute them into the plane's equation \( 2x - y + 3z = 6 \).
2. Simplify: This substitution yields the equation \( 2 - 2t + 3 + 3t = 6 \).
3. Solve: Combine like terms to simplify the equation to \( 5 - 2t = 6 \). Solve for \( t \) by first moving \( 5 \) to the other side: \(-2t = 1 \). Then, divide by \(-2\): \( t = -\frac{1}{2} \).
Finding \( t \) gives us the clue we need to pinpoint the actual coordinates where the line crosses into the plane.

Coordinate Geometry

Coordinate geometry, often called analytic geometry, is a central tool in mathematics that binds algebra and geometry. It's about using coordinate systems to explore geometric problems. This mathematical bridge allows us to find and describe the position of a point in space through numerical coordinates.
In our scenario, we've been locating the point where a line intersects a plane. The coordinates of this point are derived using the parameter \( t \) determined earlier.

• We find \( x \) using \( x = 1 - t \).
• We find \( y \) using \( y = 3t \).
• We find \( z \) using \( z = 1 + t \).

Plugging \( t = -\frac{1}{2} \) into these equations, we find the intersection point to be \( \left( \frac{3}{2}, -\frac{3}{2}, \frac{1}{2} \right) \).
Coordinate geometry not only helps in visualizing and solving such geometric problems but also aids in numerous other fields like physics, engineering, and computer science.