Problem 25 - 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 line does not intersect the plane; it is parallel.

Step 1: Write Parametric Equations of the Line

The line is given by the vector equation \ \( \langle 1,2,3 \rangle + t \langle 3,5,-1 \rangle \). We can express this as parametric equations for \(x\), \(y\), and \(z\): \[ x = 1 + 3t \]\[ y = 2 + 5t \]\[ z = 3 - t \]

Step 2: Substitute the Parametric Equations into the Plane Equation

Substitute the values of \(x\), \(y\), and \(z\) from the parametric equations into the plane equation \(3x - 2y - z = 4\).Substituting, we get: \[ 3(1 + 3t) - 2(2 + 5t) - (3 - t) = 4 \] This simplifies to \[ 3 + 9t - 4 - 10t - 3 + t = 4 \]

Step 3: Solve for the Parameter \(t\)

Simplify the equation from Step 2 to find \(t\):Combine like terms:\[ 9t - 10t + t = 4 + 4 \]\[ 0 = 5 \]This implies there is no solution for \(t\).

Step 4: Conclusion on Intersection

Since there is no solution for \(t\), the line does not intersect the plane. This implies the line is parallel to the plane.

Key Concepts

Parametric Equations and Lines

Parametric equations offer a versatile way to express the relationship of geometric lines in a spatial context. A line's path through space can be broken down into equations for each coordinate, typically expressed in terms of a parameter, often denoted as \(t\). This parameter helps trace the points along the line by varying over a range of values.

• Each equation can be visualized as describing how far along the line's direction vector each point lies, starting from a given initial point.
• For example, the line given by the vector equation \( \langle 1,2,3 \rangle + t \langle 3,5,-1 \rangle \) is represented by parametric equations: \(x = 1 + 3t\), \(y = 2 + 5t\), and \(z = 3 - t\).

This setup allows for easy substitution into other geometric forms like planes, which we explore more in the next section.

Vector Equation of a Line

The vector equation of a line combines both direction and position vector to describe the location of every point on a line.

• The fixed point, expressed as a position vector, marks the starting point of the line.
• The line's direction vector indicates the line's orientation in space.

In our example, the vector equation \( \langle 1,2,3 \rangle + t \langle 3,5,-1 \rangle \) describes a line that starts at the point \( (1, 2, 3) \) and extends infinitely in the direction of the vector \( (3, 5, -1) \). By modifying the parameter \( t \), you can explore any point along the line.
This equation captures every possible location the line might occupy as you adjust \( t \) in both positive and negative directions.

Solving Plane Equations to Find Intersections

Finding where a line and plane intersect involves substituting the line's parametric expressions into the plane's equation. The plane equation of the format \( Ax + By + Cz = D \) offers a flat surface, described in three-dimensional space.

• Integration of parametric variables from the line into the equation of the plane reveals potential intersection points.
• In our situation, substituting \(x = 1 + 3t\), \(y = 2 + 5t\), and \(z = 3 - t\) into the plane equation \(3x - 2y - z = 4\) results in an expression dependent solely on \(t\).

This resulting equation simplifies to a statement inconsistent with the original equation, indicating no valid \(t\) satisfies the equation. Therefore, the line does not physically touch the plane, suggesting parallelism, where they have the same slope but different intercepts.