Problem 26 - APEX Calculus

Question

Find the point of intersection between the line and the plane. line: \(\langle 1,2,3\rangle+t\langle 3,5,-1\rangle\) plane: \(3 x-2 y-z=-4\)

Step-by-Step Solution

Answer

The point of intersection is (1, 2, 3).

Step 1: Parameterize the Line

The equation for the line is given as \( \langle 1,2,3 \rangle + t \langle 3,5,-1 \rangle \). This indicates the line can be parameterized as:\[ x = 1 + 3t, \quad y = 2 + 5t, \quad z = 3 - t \] where \( t \) is a parameter.

Step 2: Substitute into the Plane Equation

The plane equation is \( 3x - 2y - z = -4 \). Substitute \( x = 1 + 3t \), \( y = 2 + 5t \), and \( z = 3 - t \) into the plane equation. This gives:\[ 3(1 + 3t) - 2(2 + 5t) - (3 - t) = -4 \] Simplify the expression step-by-step.

Step 3: Simplify and Solve for t

Expanding the substituted equation:\[ 3 + 9t - 4 - 10t - 3 + t = -4 \]Simplify to get:\[ 0 - t - 4 = -4 \]Solve for \( t \):\[ -t = -4 + 4 \implies -t = 0 \implies t = 0 \]

Step 4: Find the Intersection Point

Substitute \( t = 0 \) back into the parameterized line equations to find the point of intersection:\[ x = 1 + 3(0) = 1 \]\[ y = 2 + 5(0) = 2 \]\[ z = 3 - 0 = 3 \]Thus, the intersection point is \( (1, 2, 3) \).

Step 5: Verify the Intersection

To ensure the calculations are correct, check if the point \( (1, 2, 3) \) satisfies the plane equation:Substitute \( x = 1, y = 2, z = 3 \) into \( 3x - 2y - z = -4 \):\[ 3(1) - 2(2) - 3 = 3 - 4 - 3 = -4 \]Since the equation is satisfied, the calculations are confirmed to be correct.

Key Concepts

Parameterization

Parameterization helps to express a line or curve in terms of a single variable, known as a parameter. This is extremely useful in geometry when you want to express complex spatial relationships in simple algebraic forms. When a line is parameterized, it reveals how each coordinate (x, y, z) changes as the parameter varies.

• Imagine a point moving along the path of the line; the parameter is like time in this scenario, marking how far along the path you've gone.
• Each component of the line's equation is expressed in terms of the parameter, rendering the abstract idea of motion into workable math equations.

In our example, the line is defined by the fixed vector \(\langle 1, 2, 3 \rangle\) and the direction vector \(\langle 3, 5, -1 \rangle\). By scaling the direction vector with a parameter \(t\), and adding it to the fixed vector, we create expressions for \(x\), \(y\), and \(z\) as functions of \(t\). These functions allow us to understand the behavior and position of the line at various values of the parameter, thereby offering a powerful way of tackling spatial problems.

Plane Equation

Plane equations describe flat, two-dimensional surfaces stretching infinitely in three-dimensional space. These equations are characterized by a formula that relates the x, y, and z coordinates, alongside constants dictating the plane's orientation and position. In general, a plane equation is written as \(ax + by + cz = d\), where \(a\), \(b\), and \(c\) are the coefficients that determine the plane's normal vector.

• The normal vector is perpendicular to the plane and indicates its orientation.
• The constant \(d\) shifts the plane within 3D space, affecting its exact position.

For our problem, the plane is defined by \(3x - 2y - z = -4\). Each term involving \(x\), \(y\), and \(z\) is multiplied by the corresponding coefficient from the plane's normal vector \(\langle 3, -2, -1 \rangle\). This setup is ideal for calculating intersections with other objects, like lines, as it establishes a fixed relationship between spatial components.

Line Parameterization

Line parameterization breaks down a line into predictable components by introducing a variable parameter. This method is particularly useful when trying to explore how a line interacts with other geometric constructs, such as planes. By parameterizing a line, each spatial coordinate of a point on the line is expressed in terms of a parameter \(t\).

• The initial part of the line's equation \(\langle 1, 2, 3 \rangle\) indicates a fixed position from which our point starts moving.
• The term with \(t\), \(t\langle 3, 5, -1 \rangle\), describes its direction and rate of movement.

Every value of \(t\) corresponds to a point on the line, making it easy to determine where the line's path intersects with other objects, such as planes. In this exercise, the coordinates of \(x = 1 + 3t\), \(y = 2 + 5t\), and \(z = 3 - t\) illustrate how the line is parameterized with respect to the variable \(t\). This parameterized expression is fundamental for understanding and solving the problem of intersections.

Substitution

Substitution is a mathematical technique used to simplify and solve equations by replacing expressions with known values or simpler forms. In the context of finding the intersection between a line and a plane, it involves substituting the parameterized line coordinates into the plane equation. This amalgamates the two geometric structures into a single equation that can be solved for the parameter \(t\).

• By substituting \(x = 1 + 3t\), \(y = 2 + 5t\), and \(z = 3 - t\) into \(3x - 2y - z = -4\), the spatial relationship between the line and the plane becomes a single-variable equation.
• This process simplifies a 3D geometric problem into an algebraic puzzle easier to solve.

After substitution, simplifying the expression gives a value of \(t\) where the line intercepts the plane. We then use this \(t\) to find the exact coordinates of intersection by plugging it back into the parameterized line equations. This efficient method of coupling substitution with parameterization exemplifies how complex geometric relationships can be unraveled with simple math techniques.