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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![**Problem Statement:**
Find the area of a triangle \(PQR\), where \(P = (6, -3, -5)\), \(Q = (0, 4, -3)\), and \(R = (-6, -1, -5)\).
**Explanation:**
To find the area of the triangle \(PQR\), where the vertices are given in three dimensions, we use the vector cross product method. The formula for the area of a triangle with vertices \(P\), \(Q\), and \(R\) is given by:
\[ \text{Area} = \frac{1}{2} \| \overrightarrow{PQ} \times \overrightarrow{PR} \| \]
1. **Calculating Vectors:**
- Vector \(\overrightarrow{PQ}\):
\[ \overrightarrow{PQ} = Q - P = (0 - 6, 4 - (-3), -3 - (-5)) = (-6, 7, 2) \]
- Vector \(\overrightarrow{PR}\):
\[ \overrightarrow{PR} = R - P = (-6 - 6, -1 - (-3), -5 - (-5)) = (-12, 2, 0) \]
2. **Cross Product of Vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\):**
- Using the determinant method to find the cross product \(\overrightarrow{PQ} \times \overrightarrow{PR}\), we get:
\[
\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-6 & 7 & 2 \\
-12 & 2 & 0
\end{vmatrix}
\]
- Expanding the determinant:
\[
= \mathbf{i}(7 \cdot 0 - 2 \cdot 2) - \mathbf{j}(-6 \cdot 0 - 2 \cdot -12) + \mathbf{k}(-6 \cdot 2 - 7 \cdot -12)
\]
\[
= \mathbf{i}(0 - 4) - \mathbf{j}(0 +](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F55d3466e-6a31-4aff-972b-e33941523058%2Fa549c851-8370-4f91-a028-49329a46fe4f%2Fbziy8ld_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the area of a triangle \(PQR\), where \(P = (6, -3, -5)\), \(Q = (0, 4, -3)\), and \(R = (-6, -1, -5)\).
**Explanation:**
To find the area of the triangle \(PQR\), where the vertices are given in three dimensions, we use the vector cross product method. The formula for the area of a triangle with vertices \(P\), \(Q\), and \(R\) is given by:
\[ \text{Area} = \frac{1}{2} \| \overrightarrow{PQ} \times \overrightarrow{PR} \| \]
1. **Calculating Vectors:**
- Vector \(\overrightarrow{PQ}\):
\[ \overrightarrow{PQ} = Q - P = (0 - 6, 4 - (-3), -3 - (-5)) = (-6, 7, 2) \]
- Vector \(\overrightarrow{PR}\):
\[ \overrightarrow{PR} = R - P = (-6 - 6, -1 - (-3), -5 - (-5)) = (-12, 2, 0) \]
2. **Cross Product of Vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\):**
- Using the determinant method to find the cross product \(\overrightarrow{PQ} \times \overrightarrow{PR}\), we get:
\[
\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-6 & 7 & 2 \\
-12 & 2 & 0
\end{vmatrix}
\]
- Expanding the determinant:
\[
= \mathbf{i}(7 \cdot 0 - 2 \cdot 2) - \mathbf{j}(-6 \cdot 0 - 2 \cdot -12) + \mathbf{k}(-6 \cdot 2 - 7 \cdot -12)
\]
\[
= \mathbf{i}(0 - 4) - \mathbf{j}(0 +
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