Problem 59 - University Calculus: Early Transcendentals

Question

find the point in which the line meets the plane. $$x=1+2 t, \quad y=1+5 t, \quad z=3 t ; \quad x+y+z=2$$

Step-by-Step Solution

Answer

The line meets the plane at the point (1, 1, 0).

Step 1: Insert Parametric Equations into Plane Equation

The parametric equations for the line are given by: \(x = 1 + 2t\), \(y = 1 + 5t\), \(z = 3t\). We need to find when these intersect with the plane given by the equation \(x + y + z = 2\). Substitute the parametric equations into the plane equation to formulate \((1 + 2t) + (1 + 5t) + 3t = 2\).

Step 2: Simplify the Equation

Combine like terms in the equation: \(1 + 2t + 1 + 5t + 3t = 2\). This simplifies to \(2 + 10t = 2\).

Step 3: Solve for the Parameter 't'

Subtract 2 from both sides to get \(10t = 0\). Divide by 10 to find \(t = 0\).

Step 4: Find the Intersection Point

Substitute \(t = 0\) back into the parametric equations: \(x = 1 + 2(0) = 1\), \(y = 1 + 5(0) = 1\), \(z = 3(0) = 0\). Thus, the intersection point is \((1, 1, 0)\).

Step 5: Verify Solution

Verify by substituting \(x = 1\), \(y = 1\), \(z = 0\) into the plane equation: \(1 + 1 + 0 = 2\), which satisfies the plane equation. Hence, our solution is correct.

Key Concepts

Parametric Equations

Parametric equations are a way to express the coordinates of the points that make up a line (or curve) in a plane or space as functions of a variable, often denoted as \(t\). In the given exercise, the parametric equations for the line are provided as:

• \(x = 1 + 2t\)
• \(y = 1 + 5t\)
• \(z = 3t\)

These equations reveal how the x, y, and z coordinates of a point change as the parameter \(t\) varies. When \(t = 0\), we obtain the point (1, 1, 0), which is the starting point of the line. As \(t\) increases or decreases, the point moves along the direction indicated by the coefficients of \(t\) in each equation.
Understanding parametric equations helps in finding where a line intersects curves or surfaces, much like solving for intersection in this exercise.

Plane Equation

A plane in three-dimensional space can be defined using a linear equation that looks like \(Ax + By + Cz = D\). In our scenario, the plane is given by the equation \(x + y + z = 2\). This equation implies that every point (x, y, z) lying on the plane has coordinates that satisfy this condition:

• Sum of x, y, and z coordinates equals 2.

Planes can be visualized as flat, two-dimensional sheets extending infinitely in all directions within a three-dimensional space. For any given point on the plane, the equation must hold true. Thus, substituting any parametric point on a line into this plane equation will reveal whether or not this point belongs to the plane, indicating an intersection.
The given example shows the method of substitution to check which line points meet this criterion.

Intersection Point

The intersection point is where the line meets the plane; and to find this point, we use the parametric and plane equations together. From the exercise, substituting the parametric equations of the line into the plane equation \(x + y + z = 2\), lets us solve for the parameter \(t\). The simplified combined equation \(2 + 10t = 2\) is critical to isolate \(t\).

Solving gives \(t = 0\), which we plug back into the parametric equations:

• \(x = 1 + 2(0) = 1\)
• \(y = 1 + 5(0) = 1\)
• \(z = 3(0) = 0\)

Thus, the intersection point of the line and plane is (1, 1, 0). Always verify by putting these coordinates into the original plane equation to ensure the work is correct, as seen when substituting to check \(1 + 1 + 0 = 2\). Understanding this process of finding an intersection ensures you can assess points of intersection for diverse geometric scenarios.