Find the cross products u xv and vxu for the vectors u = = (3,5,0) and v = (0,3, - 6). uxv = (Di+ (Di+ (D k (Simplify your answers.)
Find the cross products u xv and vxu for the vectors u = = (3,5,0) and v = (0,3, - 6). uxv = (Di+ (Di+ (D k (Simplify your answers.)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem Statement:**
Find the cross products \( \mathbf{u} \times \mathbf{v} \) and \( \mathbf{v} \times \mathbf{u} \) for the vectors \( \mathbf{u} = \langle 3, 5, 0 \rangle \) and \( \mathbf{v} = \langle 0, 3, -6 \rangle \).
---
The expression for the cross product \( \mathbf{u} \times \mathbf{v} \) is:
\[
\mathbf{u} \times \mathbf{v} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
3 & 5 & 0 \\
0 & 3 & -6 \\
\end{vmatrix}
\]
This expands to:
\[
\mathbf{u} \times \mathbf{v} = (\text{determinant of } 3 \times 3 \text{ matrix})
\]
Evaluating:
1. \( \mathbf{i} \) component: \( | 5 \quad 0 \\ 3 \quad -6 | = (5)(-6) - (0)(3) = -30 \)
2. \( \mathbf{j} \) component: \( | 3 \quad 0 \\ 0 \quad -6 | = (3)(-6) - (0)(0) = -18 \)
3. \( \mathbf{k} \) component: \( | 3 \quad 5 \\ 0 \quad 3 | = (3)(3) - (5)(0) = 9 \)
Substituting back, we get:
\[
\mathbf{u} \times \mathbf{v} = (-30) \mathbf{i} + (-18) \mathbf{j} + 9 \mathbf{k}
\]
So,
\[
\mathbf{u} \times \mathbf{v} = \langle -30, -18, 9 \rangle
\]
---
The expression for the cross product \( \mathbf{v} \times \mathbf{u} \) is the negative of \( \mathbf{u} \times \mathbf{v} \):
\[
\mathbf{v} \times \mathbf{u} = -(\mathbf{u}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca46498c-9039-4635-914d-0b10c3becf43%2F44d9128d-da9d-424a-a5fa-e9a2474ad5b2%2Falhogup_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the cross products \( \mathbf{u} \times \mathbf{v} \) and \( \mathbf{v} \times \mathbf{u} \) for the vectors \( \mathbf{u} = \langle 3, 5, 0 \rangle \) and \( \mathbf{v} = \langle 0, 3, -6 \rangle \).
---
The expression for the cross product \( \mathbf{u} \times \mathbf{v} \) is:
\[
\mathbf{u} \times \mathbf{v} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
3 & 5 & 0 \\
0 & 3 & -6 \\
\end{vmatrix}
\]
This expands to:
\[
\mathbf{u} \times \mathbf{v} = (\text{determinant of } 3 \times 3 \text{ matrix})
\]
Evaluating:
1. \( \mathbf{i} \) component: \( | 5 \quad 0 \\ 3 \quad -6 | = (5)(-6) - (0)(3) = -30 \)
2. \( \mathbf{j} \) component: \( | 3 \quad 0 \\ 0 \quad -6 | = (3)(-6) - (0)(0) = -18 \)
3. \( \mathbf{k} \) component: \( | 3 \quad 5 \\ 0 \quad 3 | = (3)(3) - (5)(0) = 9 \)
Substituting back, we get:
\[
\mathbf{u} \times \mathbf{v} = (-30) \mathbf{i} + (-18) \mathbf{j} + 9 \mathbf{k}
\]
So,
\[
\mathbf{u} \times \mathbf{v} = \langle -30, -18, 9 \rangle
\]
---
The expression for the cross product \( \mathbf{v} \times \mathbf{u} \) is the negative of \( \mathbf{u} \times \mathbf{v} \):
\[
\mathbf{v} \times \mathbf{u} = -(\mathbf{u}
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