Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 39RE
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![**Problem Statement**
Find the volume of the parallelepiped determined by the vectors \( \vec{a} = \langle 5, 2, -1 \rangle \), \( \vec{b} = \langle 0, 1, 2 \rangle \), and \( \vec{c} = \langle 2, 4, 1 \rangle \).
**Solution**
To find the volume of the parallelepiped, we will use the scalar triple product.
The volume \(V\) is given by:
\[ V = |\vec{a} \cdot (\vec{b} \times \vec{c})| \]
**Vectors Provided:**
- \( \vec{a} = \langle 5, 2, -1 \rangle \)
- \( \vec{b} = \langle 0, 1, 2 \rangle \)
- \( \vec{c} = \langle 2, 4, 1 \rangle \)
The vector cross product \( \vec{b} \times \vec{c} \) results in another vector:
\[ \vec{b} \times \vec{c} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0 & 1 & 2 \\
2 & 4 & 1
\end{vmatrix}
= \mathbf{i}(1 \cdot 1 - 2 \cdot 4) - \mathbf{j}(0 \cdot 1 - 2 \cdot 2) + \mathbf{k}(0 \cdot 4 - 1 \cdot 2) \]
Simplify the determinant:
\[ \vec{b} \times \vec{c} = \mathbf{i}(1 - 8) - \mathbf{j}(0 - 4) + \mathbf{k}(0 - 2) \]
\[ \vec{b} \times \vec{c} = \mathbf{i}(-7) - \mathbf{j}(-4) + \mathbf{k}(-2) \]
\[ \vec{b} \times \vec{c} = \langle -7, 4, -2 \rangle \]
Now, compute the dot product \(\vec{a](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F16864198-3c29-44d0-a974-0ddc23a60123%2F2566e8bf-cf5e-45e7-82fd-4d5f5c24ea70%2Fo6neyme_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement**
Find the volume of the parallelepiped determined by the vectors \( \vec{a} = \langle 5, 2, -1 \rangle \), \( \vec{b} = \langle 0, 1, 2 \rangle \), and \( \vec{c} = \langle 2, 4, 1 \rangle \).
**Solution**
To find the volume of the parallelepiped, we will use the scalar triple product.
The volume \(V\) is given by:
\[ V = |\vec{a} \cdot (\vec{b} \times \vec{c})| \]
**Vectors Provided:**
- \( \vec{a} = \langle 5, 2, -1 \rangle \)
- \( \vec{b} = \langle 0, 1, 2 \rangle \)
- \( \vec{c} = \langle 2, 4, 1 \rangle \)
The vector cross product \( \vec{b} \times \vec{c} \) results in another vector:
\[ \vec{b} \times \vec{c} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0 & 1 & 2 \\
2 & 4 & 1
\end{vmatrix}
= \mathbf{i}(1 \cdot 1 - 2 \cdot 4) - \mathbf{j}(0 \cdot 1 - 2 \cdot 2) + \mathbf{k}(0 \cdot 4 - 1 \cdot 2) \]
Simplify the determinant:
\[ \vec{b} \times \vec{c} = \mathbf{i}(1 - 8) - \mathbf{j}(0 - 4) + \mathbf{k}(0 - 2) \]
\[ \vec{b} \times \vec{c} = \mathbf{i}(-7) - \mathbf{j}(-4) + \mathbf{k}(-2) \]
\[ \vec{b} \times \vec{c} = \langle -7, 4, -2 \rangle \]
Now, compute the dot product \(\vec{a
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