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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![**Vectors and Projections**
### Problem Description
Given two vectors:
\[ \vec{a} = \langle -5, -4, 5 \rangle \]
\[ \vec{b} = \langle 3, 1, 0 \rangle \]
The task is to find the projection of vector \(\vec{b}\) onto vector \(\vec{a}\).
### Formula for Projection
The projection of vector \(\vec{b}\) onto vector \(\vec{a}\), denoted as \(\text{proj}_{\vec{a}}\vec{b}\), is given by the formula:
\[ \text{proj}_{\vec{a}} \vec{b} = \left( \frac{\vec{a} \cdot \vec{b}}{ \vec{a} \cdot \vec{a}} \right) \vec{a} \]
### Steps to Solve
1. Calculate the dot product \(\vec{a} \cdot \vec{b}\):
\[ \vec{a} \cdot \vec{b} = (-5)(3) + (-4)(1) + (5)(0) = -15 - 4 + 0 = -19 \]
2. Calculate the dot product \(\vec{a} \cdot \vec{a}\):
\[ \vec{a} \cdot \vec{a} = (-5)^2 + (-4)^2 + (5)^2 = 25 + 16 + 25 = 66 \]
3. Compute the scalar \(\frac{\vec{a} \cdot \vec{b}}{ \vec{a} \cdot \vec{a}}\):
\[ \frac{\vec{a} \cdot \vec{b}}{\vec{a} \cdot \vec{a}} = \frac{-19}{66} \]
4. Find the vector projection:
\[ \text{proj}_{\vec{a}} \vec{b} = \left( \frac{-19}{66} \right) \langle -5, -4, 5 \rangle \]
5. Multiply the scalar with each component of vector \(\vec{a}\):
\[ \text{proj}_{\vec{a}} \vec{b](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe4962ae7-1cb2-4276-8f6a-b9f851b67289%2F69a6fbaa-6464-4e31-be31-f59840474aab%2Fhnxmdxb_processed.png&w=3840&q=75)
Transcribed Image Text:**Vectors and Projections**
### Problem Description
Given two vectors:
\[ \vec{a} = \langle -5, -4, 5 \rangle \]
\[ \vec{b} = \langle 3, 1, 0 \rangle \]
The task is to find the projection of vector \(\vec{b}\) onto vector \(\vec{a}\).
### Formula for Projection
The projection of vector \(\vec{b}\) onto vector \(\vec{a}\), denoted as \(\text{proj}_{\vec{a}}\vec{b}\), is given by the formula:
\[ \text{proj}_{\vec{a}} \vec{b} = \left( \frac{\vec{a} \cdot \vec{b}}{ \vec{a} \cdot \vec{a}} \right) \vec{a} \]
### Steps to Solve
1. Calculate the dot product \(\vec{a} \cdot \vec{b}\):
\[ \vec{a} \cdot \vec{b} = (-5)(3) + (-4)(1) + (5)(0) = -15 - 4 + 0 = -19 \]
2. Calculate the dot product \(\vec{a} \cdot \vec{a}\):
\[ \vec{a} \cdot \vec{a} = (-5)^2 + (-4)^2 + (5)^2 = 25 + 16 + 25 = 66 \]
3. Compute the scalar \(\frac{\vec{a} \cdot \vec{b}}{ \vec{a} \cdot \vec{a}}\):
\[ \frac{\vec{a} \cdot \vec{b}}{\vec{a} \cdot \vec{a}} = \frac{-19}{66} \]
4. Find the vector projection:
\[ \text{proj}_{\vec{a}} \vec{b} = \left( \frac{-19}{66} \right) \langle -5, -4, 5 \rangle \]
5. Multiply the scalar with each component of vector \(\vec{a}\):
\[ \text{proj}_{\vec{a}} \vec{b
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