Chapter 12: Problem 53
In Exercises \(53-56,\) find the point in which the line meets the plane. $$ x=1-t, \quad y=3 t, \quad z=1+t ; \quad 2 x-y+3 z=6 $$
Short Answer
Expert verified
The intersection point is \( \left(\frac{3}{2}, -\frac{3}{2}, \frac{1}{2}\right) \).
Step by step solution
01
Parametrize the Line
The line is given by the parametric equations \( x = 1 - t \), \( y = 3t \), and \( z = 1 + t \). In these equations, \( t \) is the parameter that determines the position on the line.
02
Set Up the Plane Equation
The equation of the plane is given as \( 2x - y + 3z = 6 \). To find where the line intersects the plane, substitute the parametric equations of the line into this plane equation.
03
Substitute and Solve for t
Substitute \( x = 1 - t \), \( y = 3t \), and \( z = 1 + t \) into the plane's equation: \[ 2(1 - t) - 3t + 3(1 + t) = 6 \]Simplifying the left side gives:\[ 2 - 2t - 3t + 3 + 3t = 6 \]Combine like terms:\[ 5 - 2t = 6 \]Solving for \( t \) gives:\[ -2t = 1 \Rightarrow t = -\frac{1}{2} \].
04
Find Intersection Point
Substitute \( t = -\frac{1}{2} \) back into the parametric equations:\[ x = 1 - \left(-\frac{1}{2}\right) = 1 + \frac{1}{2} = \frac{3}{2} \]\[ y = 3\left(-\frac{1}{2}\right) = -\frac{3}{2} \]\[ z = 1 + \left(-\frac{1}{2}\right) = \frac{1}{2} \]The intersection point is \( \left(\frac{3}{2}, -\frac{3}{2}, \frac{1}{2}\right) \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Equations
Parametric equations are a powerful tool to describe geometric objects using parameters. In the context of lines in 3D space, they provide a way to express the coordinates of any point on the line using a variable, commonly denoted as \( t \). This variable allows you to trace the full length of the line by changing its value.
The parametric equations given in the exercise are:
The parametric equations given in the exercise are:
- \( x = 1 - t \)
- \( y = 3t \)
- \( z = 1 + t \)
Plane Equation
Plane equations characterize flat surfaces in 3D space. They are often expressed in the general form \( ax + by + cz = d \). This linear equation relates the coordinates of any point lying on the plane, adhering to a consistent ratio between \( x \), \( y \), and \( z \).
In the exercise, the plane equation is given as \( 2x - y + 3z = 6 \). Each coefficient \((2, -1, 3)\) represents the plane's orientation, which dictates how the surface extends in three dimensions. The coefficient on the right-hand side, \( 6 \), shifts this plane parallel to its original orientation in space.
To find where a line intersects this plane, substitute the parametric equations of the line into the plane's equation. This substitution involves replacing \( x \), \( y \), and \( z \) with their parametric equivalents, simplifying the plane equation into an expression solely in terms of \( t \). This transformation reveals whether the line crosses the plane and, if so, identifies the specific \( t \) value at the intersection point.
In the exercise, the plane equation is given as \( 2x - y + 3z = 6 \). Each coefficient \((2, -1, 3)\) represents the plane's orientation, which dictates how the surface extends in three dimensions. The coefficient on the right-hand side, \( 6 \), shifts this plane parallel to its original orientation in space.
To find where a line intersects this plane, substitute the parametric equations of the line into the plane's equation. This substitution involves replacing \( x \), \( y \), and \( z \) with their parametric equivalents, simplifying the plane equation into an expression solely in terms of \( t \). This transformation reveals whether the line crosses the plane and, if so, identifies the specific \( t \) value at the intersection point.
Solving Linear Equations
Solving linear equations involves finding values for variables that satisfy a given equation. It's a critical skill for algebraic problems. In the context of intersecting a line and a plane, the process shows where the line meets the plane.
Using the exercise's context, we substitute expressions from the parametric equations into the plane equation, resulting in the equation \[2(1 - t) - 3t + 3(1 + t) = 6\]This substitution gives a single equation in one variable. The goal is to solve this linear equation for \( t \), simplifying step by step:
Using the exercise's context, we substitute expressions from the parametric equations into the plane equation, resulting in the equation \[2(1 - t) - 3t + 3(1 + t) = 6\]This substitution gives a single equation in one variable. The goal is to solve this linear equation for \( t \), simplifying step by step:
- Distribute: \( 2 - 2t - 3t + 3 + 3t = 6 \)
- Combine like terms: \( 5 - 2t = 6 \)
- Isolate \( -2t = 1 \)
- Solve for \( t \): \( t = -\frac{1}{2} \)