Problem 35 - Algebra and Trigonometry

Question

Intersection of a Line and a Plane A line has parametric equations $$ x=2+t, \quad y=3 t, \quad z=5-t $$ and a plane has equation \(5 x-2 y-2 z=1\) (a) For what value of \(t\) does the corresponding point on the line intersect the plane? (b) At what point do the line and the plane intersect?

Step-by-Step Solution

Answer

(a) The value of \(t\) is 1. (b) The intersection point is \((3, 3, 4)\).

Step 1: Write the line's parametric equations

The line's parametric equations are given as \( x = 2 + t \), \( y = 3t \), and \( z = 5 - t \). These equations describe points \((x, y, z)\) on the line in terms of the parameter \( t \).

Step 2: Substitute parametric equations into the plane's equation

Substitute \( x = 2 + t \), \( y = 3t \), and \( z = 5 - t \) into the plane equation \( 5x - 2y - 2z = 1 \). This gives: \[ 5(2 + t) - 2(3t) - 2(5 - t) = 1. \]

Step 3: Simplify the equation

Simplify the equation: \[ 10 + 5t - 6t - 10 + 2t = 1. \] Combining like terms, we get \( t = 1 \).

Step 4: Find the intersection point using \(t\)

Substitute \( t = 1 \) into the parametric equations: - \( x = 2 + 1 = 3 \) - \( y = 3 \times 1 = 3 \) - \( z = 5 - 1 = 4 \) So, the intersection point is \((3, 3, 4)\).

Step 5: Verify the intersection point

Substitute \( x = 3 \), \( y = 3 \), and \( z = 4 \) back into the plane's equation: \[ 5(3) - 2(3) - 2(4) = 15 - 6 - 8 = 1. \] The equation holds true, verifying that \((3, 3, 4)\) is indeed on the plane.

Key Concepts

Parametric Equations

Parametric equations allow us to describe a line using a parameter, often denoted by the letter \( t \). They provide a flexible way to express the coordinates of points on the line in terms of \( t \). For the line given in our exercise, the parametric equations are:

• \( x = 2 + t \)
• \( y = 3t \)
• \( z = 5 - t \)

These equations mean that you can find any point on the line by substituting a value for \( t \). Each choice of \( t \) will yield a unique point \((x, y, z)\) on the line. If \( t \) changes, the point moves along the line, helping us understand it as part of a continuous path through space.
Understanding parametric equations is crucial as they are widely used in physics, computer graphics, and engineering to model motion and pathways.

Plane Equation

A plane in three-dimensional space can be mathematically described using a plane equation. The general form of a plane equation is \( ax + by + cz = d \), where \( a, b, \) and \( c \) are constants that determine the plane's orientation, and \( d \) shifts the plane's position. In our problem, the plane equation is given by:\[ 5x - 2y - 2z = 1 \]Here, 5, -2, and -2 are the coefficients that specify the plane's orientation relative to the coordinate axes. This equation defines all points \((x, y, z)\) that lie on the plane. By assigning specific values to \(x\), \(y\), and \(z\) that satisfy this equation, we can determine whether a point is on the plane or not.
Plane equations are essential in geometry as they provide a precise mathematical way to represent flat surfaces.

Intersection Point Calculation

Finding the point where a line intersects a plane involves calculating the intersection point. This involves finding the specific value of the parameter \( t \) for which the point on the line also satisfies the plane's equation. In the original exercise, we substitute the parametric equations of the line into the plane equation:\[ 5(2 + t) - 2(3t) - 2(5 - t) = 1 \]Simplifying the left side and solving for \( t \), we find \( t = 1 \). At this value, both the line and the plane equations are satisfied.
Once \( t \) is known, the actual intersection point can be recalculated using the parametric equations.

• \( x = 2 + 1 = 3 \)
• \( y = 3 \times 1 = 3 \)
• \( z = 5 - 1 = 4 \)

Thus, the intersection point is \((3, 3, 4)\). This process is fundamental in 3D modeling and computer graphics for detecting collisions and intersections.

Algebraic Substitution

To find the intersection of a line and a plane, a key step is algebraic substitution, which involves substituting one set of equations into another. Here, the parametric equations \(x = 2 + t\), \(y = 3t\), and \(z = 5 - t\) are substituted into the plane equation \(5x - 2y - 2z = 1\). The substitution transforms the plane equation from terms of \((x, y, z)\) into terms of \(t\):\[ 5(2 + t) - 2(3t) - 2(5 - t) = 1 \]By doing this, we create an equation with only one unknown, \( t \), which simplifies the problem to normal algebra.

• Simplify the equation: combining like terms results in \(t = 1\).
• This value is then used to determine the intersection point in three-dimensional space.

Algebraic substitution is a powerful method for handling intersection problems and is widely used across mathematical fields.

Geometric Interpretation

The geometric interpretation of the intersection of a line and a plane provides visual insight into their spatial relationship. When a line intersects a plane, it does so at a single point unless it is completely contained within the plane. In this exercise, the line described by the parametric equations meets the plane at a point we calculated to be \((3, 3, 4)\).
Visualizing:

• The line as a path through space.
• The plane as a flat surface cutting through that path.
• The intersection point as where the path touches the surface.

This understanding is crucial in fields like computer graphics where rendering realistic scenes depends on correctly calculating these interactions. It helps to think of a line piercing through a sheet of paper at just one spot, providing a clear picture of intersection in 3D space.