Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![### Finding the Equation of a Plane using the Cross Product
Use the cross product to find an equation for the plane that passes through the points \( (1, 2, 3) \), \( (4, 5, 6) \), and \( (7, 8, 9) \).
**Procedure:**
1. **Identify Points:**
- Point A: \( (1, 2, 3) \)
- Point B: \( (4, 5, 6) \)
- Point C: \( (7, 8, 9) \)
2. **Find Vectors:**
- Vector AB: \( \vec{AB} = B - A = (4-1, 5-2, 6-3) = (3, 3, 3) \)
- Vector AC: \( \vec{AC} = C - A = (7-1, 8-2, 9-3) = (6, 6, 6) \)
3. **Compute the Cross Product:**
- \( \vec{n} = \vec{AB} \times \vec{AC} \)
Using the determinant method to find the cross product:
\[
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
3 & 3 & 3 \\
6 & 6 & 6
\end{vmatrix}
\]
\[
\vec{n} = (3*6 - 3*6)\mathbf{i} - (3*6 - 3*6)\mathbf{j} + (3*6 - 3*6)\mathbf{k}
\]
\[
\vec{n} = (0, 0, 0)
\]
Since the cross product result is the zero vector \( (0, 0, 0) \), it indicates that the points \( (1, 2, 3) \), \( (4, 5, 6) \), and \( (7, 8, 9) \) are collinear, and thus cannot define a unique plane.
4. **Conclusion:**
- The points lie on a straight line, so it is not possible to](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F230d8563-210a-42c1-ba8c-cbe0d0551b01%2F81eabc92-23bf-4c72-a893-856186d6d812%2Faj0zkfq_processed.png&w=3840&q=75)
Transcribed Image Text:### Finding the Equation of a Plane using the Cross Product
Use the cross product to find an equation for the plane that passes through the points \( (1, 2, 3) \), \( (4, 5, 6) \), and \( (7, 8, 9) \).
**Procedure:**
1. **Identify Points:**
- Point A: \( (1, 2, 3) \)
- Point B: \( (4, 5, 6) \)
- Point C: \( (7, 8, 9) \)
2. **Find Vectors:**
- Vector AB: \( \vec{AB} = B - A = (4-1, 5-2, 6-3) = (3, 3, 3) \)
- Vector AC: \( \vec{AC} = C - A = (7-1, 8-2, 9-3) = (6, 6, 6) \)
3. **Compute the Cross Product:**
- \( \vec{n} = \vec{AB} \times \vec{AC} \)
Using the determinant method to find the cross product:
\[
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
3 & 3 & 3 \\
6 & 6 & 6
\end{vmatrix}
\]
\[
\vec{n} = (3*6 - 3*6)\mathbf{i} - (3*6 - 3*6)\mathbf{j} + (3*6 - 3*6)\mathbf{k}
\]
\[
\vec{n} = (0, 0, 0)
\]
Since the cross product result is the zero vector \( (0, 0, 0) \), it indicates that the points \( (1, 2, 3) \), \( (4, 5, 6) \), and \( (7, 8, 9) \) are collinear, and thus cannot define a unique plane.
4. **Conclusion:**
- The points lie on a straight line, so it is not possible to
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