Chapter 12: Problem 14
Find an equation of the plane that passes through the point \(P_{0}\) with a normal vector \(\mathbf{n}\). $$P_{0}(1,2,-3) ; \mathbf{n}=\langle-1,4,-3\rangle$$
Short Answer
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Question: Find the equation of a plane that passes through point \(P_{0}(1,2,-3)\) and has a normal vector \(\mathbf{n} = \langle-1, 4, -3\rangle\).
Answer: The equation of the plane is given by \(-x + 4y - 3z = 16\).
Step by step solution
01
Convert Point \(P_{0}\) and Normal Vector \(\mathbf{n}\) to Position Vectors
We are given the point \(P_{0} (1, 2, -3)\) and the normal vector \(\mathbf{n} = \langle-1, 4, -3\rangle\). We can convert these into position vectors by writing each as a column matrix:
$$\mathbf{p_0} = \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}, \quad \mathbf{n} = \begin{pmatrix} -1 \\ 4 \\ -3 \end{pmatrix}$$
02
Write the Formula for the Equation of a Plane
Recall the general formula for the equation of a plane, given a point and a normal vector:
$$(\mathbf{n} \cdot \mathbf{r}) = (\mathbf{n} \cdot \mathbf{r_0})$$
Here, \(\mathbf{r}\) is a vector representing any point on the plane, and we already have position vectors for \(\mathbf{r_0}\) and the normal vector \(\mathbf{n}\).
03
Compute the Dot Product \((\mathbf{n} \cdot \mathbf{r_0})\)
Now, find the dot product of the normal vector \(\mathbf{n}\) and the position vector \(\mathbf{r_0}\):
$$(\mathbf{n} \cdot \mathbf{r_0}) = \begin{pmatrix} -1 \\ 4 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix} = (-1)(1) + (4)(2) + (-3)(-3) = -1 + 8 + 9 = 16$$
04
Compute the Dot Product \((\mathbf{n} \cdot \mathbf{r})\) and Find the Equation of the Plane
Now, compute the dot product of the normal vector \(\mathbf{n}\) and the vector \(\mathbf{r}\), which represents any point (x, y, z) on the plane:
$$(\mathbf{n} \cdot \mathbf{r}) = \begin{pmatrix} -1 \\ 4 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} = -x + 4y - 3z$$
Since \((\mathbf{n} \cdot \mathbf{r}) = (\mathbf{n} \cdot \mathbf{r_0})\), we can equate the two expressions we found:
$$-x + 4y - 3z = 16$$
This is the final equation of the plane passing through the point \(P_{0}(1, 2, -3)\) and having a normal vector \(\mathbf{n} = \langle-1, 4, -3\rangle\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Normal Vector
In the realm of geometry, particularly in three-dimensional space, the concept of a normal vector is crucial for understanding planes. A normal vector is a vector that is perpendicular to a given surface or plane.
This means it "sticks out" at a right angle from the surface. For a plane in 3D space, the normal vector provides essential information about the plane’s orientation.
This means it "sticks out" at a right angle from the surface. For a plane in 3D space, the normal vector provides essential information about the plane’s orientation.
- The normal vector is often denoted as \(\mathbf{n}\) and is usually expressed in component form, such as \(\langle a, b, c \rangle\).
- In the equation of a plane \(-x + 4y - 3z = 16\), the coefficients of \(x\), \(y\), and \(z\) directly come from the components of the normal vector \(\mathbf{n} = \langle -1, 4, -3 \rangle\).
- This vector provides the "directional" information of the plane, indicating how the plane is oriented in the three-dimensional space.
Dot Product
The dot product, also known as the scalar product, is a fundamental operation in vector mathematics. When you compute the dot product of two vectors, you get a scalar (a single number) rather than another vector. This operation plays a vital role in finding the equation of a plane.
- The dot product is calculated by multiplying corresponding components of the two vectors and then summing the results.
- For example, to find \(\mathbf{n} \cdot \mathbf{r_0}\), where \(\mathbf{n} = \langle -1, 4, -3 \rangle\) and \(\mathbf{r_0} = \langle 1, 2, -3 \rangle\), we do the following computation: \((-1)(1) + (4)(2) + (-3)(-3) = 16\).
Position Vectors
Position vectors are an essential concept in vector algebra, often used to locate points in space relative to a fixed origin. These vectors provide a clear way to represent any point in a three-dimensional environment.
- A position vector extends from the origin \((0,0,0)\) to the specific point.
- For instance, the position vector for point \(P_0\) (1, 2, -3) is represented as \(\begin{pmatrix} 1 \ 2 \ -3 \end{pmatrix}\).
- Position vectors help express the location of a point in terms of direction and magnitude.