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Chapter 18: Problem 3

Find an equation of the plane containing the given point \(P\) and having the given vector \(N\) as a normal vector. $$ P(2,1,-1) ; \mathbf{N}=-\mathbf{i}+3 \mathbf{j}+4 \mathbf{k} $$

Short Answer

Expert verified
The plane equation is \[-x + 3y + 4z + 3 = 0\].

Step by step solution

01

Identify Key Components

Start by identifying the given point and the components of the normal vector. The point is given as: \(P(2,1,-1)\) and the normal vector is: \(\mathbf{N} = -\mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k}\).
02

Convert Normal Vector to Components

Write the normal vector in component form: \(\mathbf{N} = -1\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}\). This translates to the vector components \(A = -1\), \(B = 3\), and \(C = 4\).
03

Use the Point-Normal Form of Plane Equation

The equation of a plane given a point \((x_0, y_0, z_0)\) and a normal vector \(\langle A, B, C \rangle\) is: \(A(x - x_0) + B(y - y_0) + C(z - z_0) = 0\). Substitute \(A, B, C\) and \(P(2,1,-1)\) into the equation.
04

Substitute Values into Equation

Given \(A = -1\), \(B = 3\), and \(C = 4\), and point \(P(2, 1, -1)\), substitute into the plane equation: \[-1(x - 2) + 3(y - 1) + 4(z + 1) = 0\].
05

Simplify the Equation

Expand and simplify the equation: \[-1(x - 2) + 3(y - 1) + 4(z + 1) = 0\] \[-x + 2 + 3y - 3 + 4z + 4 = 0\] Combine like terms: \[-x + 3y + 4z + 3 = 0\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Vector
A normal vector is a vector that is perpendicular to a surface or plane. In the context of the problem, the normal vector is \(\textbf{N} = -\textbf{i} + 3\textbf{j} + 4\textbf{k}\). This vector signifies the direction orthogonal (at right angles) to the plane we are looking to define.

To understand why the normal vector is important:
  • It helps determine the plane's orientation in space.
  • It's crucial for calculating the plane equation since the plane will be orthogonal to this vector.
The components of this normal vector influence the slope and position of the resultant plane.
Point-Normal Form
The Point-Normal Form is a way to represent the equation of a plane when you have a known point on the plane and a normal vector that is perpendicular to the plane.

This form of the equation is given by:
\[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \]
where:
  • \( \text{(x}_0, \text{y}_0, \text{z}_0) \) are the coordinates of a known point on the plane.
  • \( A, B, \text{and} C \) are the components of the normal vector.
In our example, the known point is \( P(2, 1, -1) \) and the normal vector components are \( A = -1, B = 3, \text{and} C = 4 \). Using the point-normal form allows us to easily substitute these values and derive the plane's equation.
Vector Components
Vector components break down a vector into its constituent parts along the coordinate axes.

For the normal vector \( \textbf{N} = -\textbf{i} + 3\textbf{j} + 4\textbf{k} \), the components are:
  • \( A = -1 \) along the x-axis.
  • \( B = 3 \) along the y-axis.
  • \( C = 4 \) along the z-axis.
By converting the vector into component form, we retrieve the essential coefficients needed to define the plane equation. These components dictate how the plane is slanted in three-dimensional space relative to each axis.
Plane Equation
To find the equation of a plane, we integrate the normal vector components and a given point into the point-normal form equation.

From the exercise, we have:
  • Point: \( P(2, 1, -1) \).
  • Normal vector components: \( A = -1, B = 3, \text{and} C = 4 \).
Using the point-normal form, we substitute these into the equation \( A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \):
\[ -1(x - 2) + 3(y - 1) + 4(z + 1) = 0 \]
Next, we simplify:
\[ -x + 2 + 3y - 3 + 4z + 4 = 0 \]
Combining like terms, we get:
\[ -x + 3y + 4z + 3 = 0 \]
Thus, the equation representing our plane is:
\[ -x + 3y + 4z + 3 = 0 \].
This final form is the standard equation of the plane, integrating all the given conditions.

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Most popular questions from this chapter

Find an equation in cylindrical coordinates of the given surface and identify the surface. $$ x^{2}-y^{2}=3 z^{2} $$

Draw a sketch of the graph of the given equation and name the surface. $$ 4 x^{2}+9 y^{2}+z^{2}=36 $$

Find an equation of the surface of revolution generated by revolving the given plane curve about the indicated axis. Draw a sketch of the surface. \(x^{2}+4 z^{2}=16\) in the \(x z\) plane, about the \(z\) axis.

Show that the lines $$ \frac{x+1}{2}=\frac{y+4}{-5}=\frac{z-2}{3} \text { and } \frac{x-3}{-2}=\frac{y+14}{5}=\frac{z-8}{-3} $$ are coincident.

Find an equation of the surface of revolution generated by revolving the given plane curve about the indicated axis. Draw a sketch of the surface. \(x^{2}+4 z^{2}=16\) in the \(x z\) plane, about the \(x\) axis.
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