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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![### Cross Product Calculation with Integer Constraints
#### Problem Statement
The blanks can only take integers.
What is the cross product of \( \langle 2, 1, 0 \rangle \) and \( \langle 3, -2, 4 \rangle \)?
#### Explanation
To find the cross product of two vectors, \(\mathbf{A}\) and \(\mathbf{B}\), in three-dimensional space, we use the following formula:
\[
\mathbf{A} \times \mathbf{B} = \langle a_1, a_2, a_3 \rangle \times \langle b_1, b_2, b_3 \rangle = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle
\]
Substituting the given vectors into the formula, we get:
\[
\langle 2, 1, 0 \rangle \times \langle 3, -2, 4 \rangle
\]
Let's denote:
\[
a_1 = 2, a_2 = 1, a_3 = 0
\]
\[
b_1 = 3, b_2 = -2, b_3 = 4
\]
Now, calculate each component of the resulting vector:
\[
\left| \begin{matrix}
a_2 & a_3 \\
b_2 & b_3 \\
\end{matrix} \right| = \left| \begin{matrix}
1 & 0 \\
-2 & 4 \\
\end{matrix} \right| = (1 \cdot 4) - (0 \cdot -2) = 4
\]
\[
\left| \begin{matrix}
a_3 & a_1 \\
b_3 & b_1 \\
\end{matrix} \right| = \left| \begin{matrix}
0 & 2 \\
4 & 3 \\
\end{matrix} \right| = (0 \cdot 3) - (2 \cdot 4) = -8
\]
\[
\left| \begin{matrix}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdfe80350-6fa1-469d-aae2-95b4dd3aa499%2F96a26d70-1a69-48e6-9761-0db93fdbc8ee%2Fwc3ips_processed.png&w=3840&q=75)
Transcribed Image Text:### Cross Product Calculation with Integer Constraints
#### Problem Statement
The blanks can only take integers.
What is the cross product of \( \langle 2, 1, 0 \rangle \) and \( \langle 3, -2, 4 \rangle \)?
#### Explanation
To find the cross product of two vectors, \(\mathbf{A}\) and \(\mathbf{B}\), in three-dimensional space, we use the following formula:
\[
\mathbf{A} \times \mathbf{B} = \langle a_1, a_2, a_3 \rangle \times \langle b_1, b_2, b_3 \rangle = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle
\]
Substituting the given vectors into the formula, we get:
\[
\langle 2, 1, 0 \rangle \times \langle 3, -2, 4 \rangle
\]
Let's denote:
\[
a_1 = 2, a_2 = 1, a_3 = 0
\]
\[
b_1 = 3, b_2 = -2, b_3 = 4
\]
Now, calculate each component of the resulting vector:
\[
\left| \begin{matrix}
a_2 & a_3 \\
b_2 & b_3 \\
\end{matrix} \right| = \left| \begin{matrix}
1 & 0 \\
-2 & 4 \\
\end{matrix} \right| = (1 \cdot 4) - (0 \cdot -2) = 4
\]
\[
\left| \begin{matrix}
a_3 & a_1 \\
b_3 & b_1 \\
\end{matrix} \right| = \left| \begin{matrix}
0 & 2 \\
4 & 3 \\
\end{matrix} \right| = (0 \cdot 3) - (2 \cdot 4) = -8
\]
\[
\left| \begin{matrix}
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