Calculus: Early Transcendentals
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ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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![**Finding the Plane Equation**
**Problem:**
Determine the equation of a plane that passes through the point (6, 7, -2) and contains the line given by the parametric equations:
\[ x = 4 - 2t, \ y = 3 + 8t, \ z = 7 + 4t \]
**Solution:**
1. **Given Data:**
- Point: \((6, 7, -2)\)
- Line: \(x = 4 - 2t, \ y = 3 + 8t, \ z = 7 + 4t\)
2. **Direction Vector of the Line:**
The parametric equations can be written as:
\[
\vec{r} = \langle 4, 3, 7 \rangle + t \langle -2, 8, 4 \rangle
\]
The direction vector is:
\[
\vec{d} = \langle -2, 8, 4 \rangle
\]
3. **Vector from Point on the Line to the Given Point:**
Let \( P = (6, 7, -2) \) be the given point and \( Q = (4, 3, 7) \) be a point on the line when \( t = 0 \).
Vector \( \vec{PQ} = \langle 6 - 4, 7 - 3, -2 - 7 \rangle = \langle 2, 4, -9 \rangle \)
4. **Normal Vector to the Plane (Cross Product):**
Compute the normal vector \( \vec{n} \) to the plane which is perpendicular to both \( \vec{PQ} \) and the direction vector \( \vec{d} \):
\[
\vec{n} = \vec{PQ} \times \vec{d}
\]
Calculate the cross product:
\[
\vec{n} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 4 & -9 \\
-2 & 8 & 4 \\
\end{vmatrix}
\]
\](https://content.bartleby.com/qna-images/question/7bb6f29c-bae7-4b63-bf8e-21de0c88ea98/d6677e9f-575a-4e07-971a-9503078f5dfb/kv60ca.jpeg)
Transcribed Image Text:**Finding the Plane Equation**
**Problem:**
Determine the equation of a plane that passes through the point (6, 7, -2) and contains the line given by the parametric equations:
\[ x = 4 - 2t, \ y = 3 + 8t, \ z = 7 + 4t \]
**Solution:**
1. **Given Data:**
- Point: \((6, 7, -2)\)
- Line: \(x = 4 - 2t, \ y = 3 + 8t, \ z = 7 + 4t\)
2. **Direction Vector of the Line:**
The parametric equations can be written as:
\[
\vec{r} = \langle 4, 3, 7 \rangle + t \langle -2, 8, 4 \rangle
\]
The direction vector is:
\[
\vec{d} = \langle -2, 8, 4 \rangle
\]
3. **Vector from Point on the Line to the Given Point:**
Let \( P = (6, 7, -2) \) be the given point and \( Q = (4, 3, 7) \) be a point on the line when \( t = 0 \).
Vector \( \vec{PQ} = \langle 6 - 4, 7 - 3, -2 - 7 \rangle = \langle 2, 4, -9 \rangle \)
4. **Normal Vector to the Plane (Cross Product):**
Compute the normal vector \( \vec{n} \) to the plane which is perpendicular to both \( \vec{PQ} \) and the direction vector \( \vec{d} \):
\[
\vec{n} = \vec{PQ} \times \vec{d}
\]
Calculate the cross product:
\[
\vec{n} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 4 & -9 \\
-2 & 8 & 4 \\
\end{vmatrix}
\]
\
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