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...
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![**Problem 8:**
Given the points \( P(3, -9) \) and \( Q(0, -3) \), find the component form of the vector \( \overrightarrow{PQ} \) and its magnitude \( |\overrightarrow{PQ}| \).
### Explanation:
To find the component form of the vector \( \overrightarrow{PQ} \), subtract the coordinates of point \( P \) from point \( Q \):
- \( x \)-component: \( 0 - 3 = -3 \)
- \( y \)-component: \( -3 - (-9) = 6 \)
Thus, the component form of \( \overrightarrow{PQ} \) is \((-3, 6)\).
To find the magnitude \( |\overrightarrow{PQ}| \), use the formula for the magnitude of a vector:
\[
|\overrightarrow{PQ}| = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9f16a484-99a6-4311-bd1f-aad436adf7e1%2Ff64b6eda-cec2-425d-be61-69ed58634d8f%2Fir0t70i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 8:**
Given the points \( P(3, -9) \) and \( Q(0, -3) \), find the component form of the vector \( \overrightarrow{PQ} \) and its magnitude \( |\overrightarrow{PQ}| \).
### Explanation:
To find the component form of the vector \( \overrightarrow{PQ} \), subtract the coordinates of point \( P \) from point \( Q \):
- \( x \)-component: \( 0 - 3 = -3 \)
- \( y \)-component: \( -3 - (-9) = 6 \)
Thus, the component form of \( \overrightarrow{PQ} \) is \((-3, 6)\).
To find the magnitude \( |\overrightarrow{PQ}| \), use the formula for the magnitude of a vector:
\[
|\overrightarrow{PQ}| = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}
\]
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