Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Related questions
Question
![### Problem:
Find the distance from the point \((-5, 5, -6)\) to the line \(\vec{a}(t) = \langle 5, 1, -4 \rangle t + \langle 2, -1, 0 \rangle\).
### Solution:
To find the distance between a point and a line in three-dimensional space, we can use vector calculus methods. The distance \( D \) from a point \( P \) to a line with vector equation \(\vec{r}(t) = \vec{a} t + \vec{b} \) can be found using the following steps:
1. Let \( \vec{P} \) be the position vector of the point.
2. Let \( \vec{r_0} = \vec{b} \) be any point on the line.
3. Let \( \vec{d} = \vec{a} \) be the direction vector of the line.
4. The desired distance \( D \) is given by:
\[ D = \frac{||(\vec{P} - \vec{r_0}) \times \vec{d}||}{||\vec{d}||} \]
Where:
- \( \vec{P} \) is the position vector of the point \((-5, 5, -6)\).
- \( \vec{r_0} \) is a point on the line \( \langle 2, -1, 0 \rangle \).
- \( \vec{d} \) is the direction vector of the line \( \langle 5, 1, -4 \rangle \).
By substituting these values into the formula, we can compute the distance.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F16864198-3c29-44d0-a974-0ddc23a60123%2F3ade6652-2cc2-41fa-be74-b2a2790f7fd0%2Fg94ofle_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem:
Find the distance from the point \((-5, 5, -6)\) to the line \(\vec{a}(t) = \langle 5, 1, -4 \rangle t + \langle 2, -1, 0 \rangle\).
### Solution:
To find the distance between a point and a line in three-dimensional space, we can use vector calculus methods. The distance \( D \) from a point \( P \) to a line with vector equation \(\vec{r}(t) = \vec{a} t + \vec{b} \) can be found using the following steps:
1. Let \( \vec{P} \) be the position vector of the point.
2. Let \( \vec{r_0} = \vec{b} \) be any point on the line.
3. Let \( \vec{d} = \vec{a} \) be the direction vector of the line.
4. The desired distance \( D \) is given by:
\[ D = \frac{||(\vec{P} - \vec{r_0}) \times \vec{d}||}{||\vec{d}||} \]
Where:
- \( \vec{P} \) is the position vector of the point \((-5, 5, -6)\).
- \( \vec{r_0} \) is a point on the line \( \langle 2, -1, 0 \rangle \).
- \( \vec{d} \) is the direction vector of the line \( \langle 5, 1, -4 \rangle \).
By substituting these values into the formula, we can compute the distance.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 3 images
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning