Chapter 14: Problem 6
Find the equation of a line through (-2,1,4) and parallel to the vector \(3 \mathbf{i}-4 \mathbf{j}+7 \mathbf{k}\).
Short Answer
Expert verified
The line's equation is \(x = -2 + 3\lambda\), \(y = 1 - 4\lambda\), \(z = 4 + 7\lambda\).
Step by step solution
01
Identify the Direction Vector
The direction vector for the line is given as \(3\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}\). This will be used as the direction vector of the line that is parallel to it.
02
Use the Point-Vector Form of the Line Equation
A line in 3D can be described using a position vector \( extbf{r}\) and a direction vector \( extbf{d}\): \(\textbf{r} = \textbf{a} + \lambda \textbf{d}\), where \(\lambda\) is a parameter. Here, \(\textbf{a} = -2\mathbf{i} + 1\mathbf{j} + 4\mathbf{k}\) is the given point and \(\textbf{d} = 3\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}\) is the given direction vector.
03
Write the Equation of the Line
Substitute the point \((-2, 1, 4)\) and the direction vector \((3, -4, 7)\) into the equation: \(\textbf{r} = (-2)\mathbf{i} + 1\mathbf{j} + 4\mathbf{k} + \lambda(3\mathbf{i} - 4\mathbf{j} + 7\mathbf{k})\). After substituting, the parametric equations of the line are: \(x = -2 + 3\lambda\), \(y = 1 - 4\lambda\), \(z = 4 + 7\lambda\).
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.
Direction Vector
In three-dimensional space, a direction vector is fundamental for defining the orientation of a line. It essentially acts like a compass, indicating the direction in which the line is pointing.
The direction vector is usually represented by the components enclosed in angle brackets or as a combination of unit vectors such as \(\mathbf{i}, \mathbf{j}, \) and \(\mathbf{k}\).
For instance, in the given problem, the direction vector is \(3\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}.\) This vector shows us how the line moves in the x, y, and z directions.
The direction vector is usually represented by the components enclosed in angle brackets or as a combination of unit vectors such as \(\mathbf{i}, \mathbf{j}, \) and \(\mathbf{k}\).
For instance, in the given problem, the direction vector is \(3\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}.\) This vector shows us how the line moves in the x, y, and z directions.
- The first component, 3, means the line increases by 3 units along the x-axis for each step.
- Similarly, -4 indicates a decrease along the y-axis, while 7 shows an increase along the z-axis.
Point-Vector Form
The point-vector form of a line in three-dimensional space is an elegant way to represent the line using both its direction and a single known point.
This form is expressed mathematically as \(\textbf{r} = \textbf{a} + \lambda \textbf{d},\) where \(\textbf{r}\) is the position vector of any point on the line, \(\textbf{a}\) is a known point on the line, and \(\textbf{d}\) is the direction vector. The parameter \(\lambda\) is a scalar that allows one to move along the line.
This form is expressed mathematically as \(\textbf{r} = \textbf{a} + \lambda \textbf{d},\) where \(\textbf{r}\) is the position vector of any point on the line, \(\textbf{a}\) is a known point on the line, and \(\textbf{d}\) is the direction vector. The parameter \(\lambda\) is a scalar that allows one to move along the line.
- In our current example, \(\textbf{a}\) is the point \(-2\mathbf{i} + 1\mathbf{j} + 4\mathbf{k}\).
- This form combines both the direction of the line and its position in space.
Parametric Equations
Parametric equations are a set of equations that express the coordinates of the points on a line—or any curve—as a function of one or more parameters.
These provide a flexible form of representing lines, particularly in three-dimensional space, by indicating how x, y, and z coordinates change based on a parameter \(\lambda\).
Given the vector equation \(\textbf{r} = \textbf{a} + \lambda \textbf{d},\), it can be split into separate equations for each coordinate.
For our problem, substituting the given point and direction vector, we obtain the parametric equations:
These provide a flexible form of representing lines, particularly in three-dimensional space, by indicating how x, y, and z coordinates change based on a parameter \(\lambda\).
Given the vector equation \(\textbf{r} = \textbf{a} + \lambda \textbf{d},\), it can be split into separate equations for each coordinate.
For our problem, substituting the given point and direction vector, we obtain the parametric equations:
- \(x = -2 + 3\lambda\),
- \(y = 1 - 4\lambda\),
- \(z = 4 + 7\lambda\).