Problem 55 - Thomas Calculus

Question

In Exercises \(53-56,\) 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 intersects the plane at the point (1, 1, 0).

Step 1: Substitute Line Equations into Plane Equation

The equations for the line are given by:\[ x = 1 + 2t, \ y = 1 + 5t, \ z = 3t. \]These need to be substituted into the plane equation: \[ x + y + z = 2. \]Substituting, we get:\[ (1 + 2t) + (1 + 5t) + 3t = 2. \]

Step 2: Simplify and Solve for t

Combine the terms from the substitution:\[ 1 + 2t + 1 + 5t + 3t = 2. \]This simplifies to:\[ 2 + 10t = 2. \]Subtract 2 from both sides:\[ 10t = 0. \]Divide by 10:\[ t = 0. \]

Step 3: Find the Point of Intersection

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

Step 4: Write the Coordinates of the Intersection Point

The coordinates of the point where the line intersects the plane are \( (1, 1, 0) \).

Key Concepts

Equations of Lines

Understanding the equations of lines helps us navigate the world of geometry and algebra. Lines in three-dimensional space can be represented parametrically, meaning they involve a parameter like "t" to describe a set of points along the line. Each equation specifies how coordinates change in relation to the parameter.
In the example exercise, the line is expressed through three equations:

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

Here, as "t" changes, different points on the line are described. Each equation describes how much the corresponding coordinate of a point on the line increases from a particular starting point. This parametrization makes it easier to calculate intersections with planes by adjusting the parameter as needed.

Equations of Planes

Planes, unlike lines, extend infinitely in two directions in three-dimensional space. A common way to describe a plane is through a linear equation in the form \( ax + by + cz = d \), where \(a\), \(b\), and \(c\) are constants that define the plane's orientation, and \(d\) gives the position relative to the origin.
In our exercise, the plane equation is:

• \( x + y + z = 2 \)

This equation suggests that any point \((x, y, z)\) lying on this plane will satisfy the equation when substituted in place of \(x\), \(y\), and \(z\). In solving the intersection problem, we substitute the line's parametric equations into this plane equation. By solving the resulting equation, we can determine values for "t" that locate points common to both the line and the plane.

Linear Algebra Concepts

Linear algebra provides powerful tools for understanding spaces and intersections, combining algebraic techniques with geometric interpretation. One essential concept is transforming equations to find intersections, like the line-plane intersection in our example.
By substituting the parametric line equations into the plane equation, we convert a geometric problem into an algebraic one.
The intersection point calculation exemplifies the usefulness of systems of equations and linear transformation principles:

• Setting up substitutable expressions (from lines) informs the simultaneous equations approach.
• Simplifying those expressions helps reach a scalar equation—here, \(10t = 0\)—yielding solutions for parameters (like "t").
• Solved parameters direct back to specific points in space once substituted again into original expressions.

Linear algebra also lays groundwork for more advanced concepts, including projections and transformations, foundational in understanding complex shapes and movements within spaces.