Chapter 11: Problem 25
Find the projection \(P\) of \(B\) along A. Also find \(|P|\). \(\mathbf{A}\) is perpendicular to \(x-y+z=0, \mathbf{B}=\mathbf{i}+\mathbf{j}\).
Short Answer
Expert verified
The projection \(P\) is the zero vector, with a magnitude of 0.
Step by step solution
01
Identify the Normal Vector
The vector \(\mathbf{A}\) is perpendicular to the plane given by the equation \(x-y+z=0\). The coefficients of \(x, y,\) and \(z\) in this equation provide the normal vector to the plane. Therefore, the normal vector is \(\mathbf{A} = \mathbf{i} - \mathbf{j} + \mathbf{k}\).
02
Calculate the Dot Product of \(\mathbf{A}\) and \(\mathbf{B}\)
Calculate the dot product \(\mathbf{A} \cdot \mathbf{B}\) where \(\mathbf{B}=\mathbf{i}+\mathbf{j}\). So, \(\mathbf{A} \cdot \mathbf{B} = (1)(1) + (-1)(1) + (1)(0) = 1 - 1 + 0 = 0\).
03
Compute the Magnitude of \(\mathbf{A}\)
Calculate the magnitude of \(\mathbf{A}\) using the formula \(||\mathbf{A}|| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}\).
04
Find the Projection of \(\mathbf{B}\) onto \(\mathbf{A}\)
The projection formula is given as \(P = \frac{\mathbf{A} \cdot \mathbf{B}}{\mathbf{A} \cdot \mathbf{A}} \mathbf{A}\). Since \(\mathbf{A} \cdot \mathbf{B} = 0\), the projection \(P\) will be \(\mathbf{0}\) (a zero vector).
05
Calculate the Magnitude of the Projection
Since the projection \(P\) is the zero vector, its magnitude is simply \(|P| = 0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Dot Product
The dot product, also known as the scalar product, is a key operation in vector algebra. This operation multiplies two vectors, resulting in a single number, also known as a scalar. To compute the dot product of two vectors, we multiply their corresponding components and add them together. For example, given vectors \(\mathbf{A} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}\) and \(\mathbf{B} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k}\), the dot product \(\mathbf{A} \cdot \mathbf{B}\) is calculated as:
- \(\mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 + a_3b_3\).
Magnitude of a Vector
The magnitude of a vector measures its length and is always a non-negative value. It essentially tells us how long the vector is in the vector space. To find the magnitude of a vector \(\mathbf{V} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k}\), we use the formula:
- \(||\mathbf{V}|| = \sqrt{a^2 + b^2 + c^2}\).
Normal Vector
A normal vector is a vector that is perpendicular to a surface. In the context of a plane given by a linear equation of the form \(ax + by + cz = d\), the normal vector is \(\mathbf{N} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k}\). This vector points directly away from the plane. For example, the equation \(x-y+z=0\) defines a plane, and its normal vector is \(\mathbf{i} - \mathbf{j} + \mathbf{k}\). Recognizing the concept of a normal vector is crucial when dealing with projections, intersections, and reflections with respect to planes. It helps determine how surfaces are oriented in three-dimensional space.
Zero Vector
The zero vector, denoted as \(\mathbf{0}\), is a vector with all its components equal to zero. In a three-dimensional space, it is represented as \(\mathbf{0} = 0 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k}\). Despite having zero magnitude, it is still considered a vector because it represents the origin or a point of no displacement in the vector space.Zero vectors often arise in calculations involving projections, especially when dealing with perpendicular vectors. In the original exercise, when projecting vector \(\mathbf{B}\) onto \(\mathbf{A}\), the result is a zero vector because \(\mathbf{A} \cdot \mathbf{B} = 0\). This indicates that there is no component of \(\mathbf{B}\) in the direction of \(\mathbf{A}\), highlighting the perpendicular nature of the vectors. Understanding zero vectors is vital, especially in linear algebra and computer graphics, where they are frequently used in transformation and optimization tasks.