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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![### Vector Cross Product Problem
#### Problem Statement:
Let \(\mathbf{u} \times \mathbf{v} = \langle 0, 3, 2 \rangle\). What is \((6 \mathbf{u} + 3 \mathbf{v}) \times \mathbf{u}\)?
#### Answer Choices:
A) \(\langle 0, 18, 12 \rangle\)
B) \(\langle 0, 0, 0 \rangle\)
C) \(\langle 0, -9, -6 \rangle\)
D) \(\langle 0, 9, 6 \rangle\)
#### Explanation:
To solve for \((6 \mathbf{u} + 3 \mathbf{v}) \times \mathbf{u}\), we can use the distributive property of the cross product:
\[ (6 \mathbf{u} + 3 \mathbf{v}) \times \mathbf{u} = 6 (\mathbf{u} \times \mathbf{u}) + 3 (\mathbf{v} \times \mathbf{u}) \]
1. Note that the cross product of any vector with itself is always a zero vector:
\[ \mathbf{u} \times \mathbf{u} = \mathbf{0} \]
2. Given \(\mathbf{u} \times \mathbf{v} = \langle 0, 3, 2 \rangle\), we can find \(\mathbf{v} \times \mathbf{u}\) by negating the result (due to the anti-commutative property of the cross product):
\[ \mathbf{v} \times \mathbf{u} = -(\mathbf{u} \times \mathbf{v}) = -\langle 0, 3, 2 \rangle = \langle 0, -3, -2 \rangle \]
3. Using these results in the equation, we get:
\[ (6 \mathbf{u} + 3 \mathbf{v}) \times \mathbf{u} = 6 \mathbf{0} + 3 \langle 0, -3, -2 \rangle = \langle 0, 0, 0 \rangle + \langle 0, -9, -6](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F64775454-4845-4a2e-95ce-edb2bc3db137%2Fb4b23f9d-0b9a-40e4-adef-f6277d70e208%2Fygqont.png&w=3840&q=75)
Transcribed Image Text:### Vector Cross Product Problem
#### Problem Statement:
Let \(\mathbf{u} \times \mathbf{v} = \langle 0, 3, 2 \rangle\). What is \((6 \mathbf{u} + 3 \mathbf{v}) \times \mathbf{u}\)?
#### Answer Choices:
A) \(\langle 0, 18, 12 \rangle\)
B) \(\langle 0, 0, 0 \rangle\)
C) \(\langle 0, -9, -6 \rangle\)
D) \(\langle 0, 9, 6 \rangle\)
#### Explanation:
To solve for \((6 \mathbf{u} + 3 \mathbf{v}) \times \mathbf{u}\), we can use the distributive property of the cross product:
\[ (6 \mathbf{u} + 3 \mathbf{v}) \times \mathbf{u} = 6 (\mathbf{u} \times \mathbf{u}) + 3 (\mathbf{v} \times \mathbf{u}) \]
1. Note that the cross product of any vector with itself is always a zero vector:
\[ \mathbf{u} \times \mathbf{u} = \mathbf{0} \]
2. Given \(\mathbf{u} \times \mathbf{v} = \langle 0, 3, 2 \rangle\), we can find \(\mathbf{v} \times \mathbf{u}\) by negating the result (due to the anti-commutative property of the cross product):
\[ \mathbf{v} \times \mathbf{u} = -(\mathbf{u} \times \mathbf{v}) = -\langle 0, 3, 2 \rangle = \langle 0, -3, -2 \rangle \]
3. Using these results in the equation, we get:
\[ (6 \mathbf{u} + 3 \mathbf{v}) \times \mathbf{u} = 6 \mathbf{0} + 3 \langle 0, -3, -2 \rangle = \langle 0, 0, 0 \rangle + \langle 0, -9, -6
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