Problem 21
Question
Decompose \(\mathbf{v}\) into two vectors \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\), where \(\mathbf{v}_{1}\) is parallel to \(\mathbf{w}\), and \(\mathbf{v}_{2}\) is orthogonal to \(\mathbf{w}\). $$ \mathbf{v}=2 \mathbf{i}-3 \mathbf{j}, \quad \mathbf{w}=\mathbf{i}-\mathbf{j} $$
Step-by-Step Solution
Verified Answer
\(\mathbf{v}_{1} = \frac{5}{2} \mathbf{i} - \frac{5}{2} \mathbf{j}\) and \(\mathbf{v}_{2} = -\frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j}\)
1Step 1: Represent the vectors \(\mathbf{v}\) and \(\mathbf{w}\)
Express the given vectors in component form. \(\mathbf{v} = 2\mathbf{i} - 3\mathbf{j}\) and \(\mathbf{w} = \mathbf{i} - \mathbf{j}\)
2Step 2: Find the projection of \(\mathbf{v}\) onto \(\mathbf{w}\)
The projection of \(\mathbf{v}\) onto \(\mathbf{w}\) is given by: \[\mathbf{v}_{1} = \frac{\mathbf{v} \cdot \mathbf{w}}{\mathbf{w} \cdot \mathbf{w}} \mathbf{w}\]. \(\mathbf{v} \cdot \mathbf{w} = (2)(1) + (-3)(-1) = 5\). Also, \(\mathbf{w} \cdot \mathbf{w} = (1)^2 + (-1)^2 = 2\). Hence, \[\mathbf{v}_{1} = \frac{5}{2} \mathbf{w} = \frac{5}{2} (\mathbf{i} - \mathbf{j}) = \frac{5}{2} \mathbf{i} - \frac{5}{2} \mathbf{j}\]
3Step 3: Determine the orthogonal component \(\mathbf{v}_{2}\)
The orthogonal component \(\mathbf{v}_{2}\) is found by subtracting the parallel component \(\mathbf{v}_{1}\) from \(\mathbf{v}\). \[\mathbf{v}_{2} = \mathbf{v} - \mathbf{v}_{1} = (2 \mathbf{i} - 3 \mathbf{j}) - (\frac{5}{2} \mathbf{i} - \frac{5}{2} \mathbf{j})\]. Simplify the expression: \[(2 - \frac{5}{2}) \mathbf{i} + (-3 + \frac{5}{2}) \mathbf{j} = -\frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j}\]
Key Concepts
Vector ProjectionOrthogonal VectorsComponent Form
Vector Projection
Vector projection is a method to project one vector onto another vector. This process helps us understand how much of one vector goes in the direction of another vector.
Let's break it down: imagine you have a vector \(\textbf{v}\) and you want to see the part of it that points in the direction of another vector \(\textbf{w}\). This part is called the projection of \(\textbf{v}\) onto \(\textbf{w}\).
To find the projection, you use the formula: \(\textbf{v}_{1} = \frac{\textbf{v} \cdot \textbf{w}}{\textbf{w} \cdot \textbf{w}} \textbf{w}\). This formula involves taking the dot product of \(\textbf{v}\) and \(\textbf{w}\), then dividing by the dot product of \(\textbf{w}\) with itself. The projection vector \(\textbf{v}_{1}\) shows the part of \(\textbf{v}\) that aligns with \(\textbf{w}\).
In the given exercise, the projection step is: \(\textbf{v}_{1} = \frac{5}{2} \textbf{w} = \frac{5}{2} (\textbf{i} - \textbf{j}) = \frac{5}{2} \textbf{i} - \frac{5}{2} \textbf{j}\). This represents the portion of \(\textbf{v}\) that is parallel to \(\textbf{w}\).
Let's break it down: imagine you have a vector \(\textbf{v}\) and you want to see the part of it that points in the direction of another vector \(\textbf{w}\). This part is called the projection of \(\textbf{v}\) onto \(\textbf{w}\).
To find the projection, you use the formula: \(\textbf{v}_{1} = \frac{\textbf{v} \cdot \textbf{w}}{\textbf{w} \cdot \textbf{w}} \textbf{w}\). This formula involves taking the dot product of \(\textbf{v}\) and \(\textbf{w}\), then dividing by the dot product of \(\textbf{w}\) with itself. The projection vector \(\textbf{v}_{1}\) shows the part of \(\textbf{v}\) that aligns with \(\textbf{w}\).
In the given exercise, the projection step is: \(\textbf{v}_{1} = \frac{5}{2} \textbf{w} = \frac{5}{2} (\textbf{i} - \textbf{j}) = \frac{5}{2} \textbf{i} - \frac{5}{2} \textbf{j}\). This represents the portion of \(\textbf{v}\) that is parallel to \(\textbf{w}\).
Orthogonal Vectors
Orthogonal vectors are vectors that are perpendicular to each other. When two vectors are orthogonal, their dot product is zero.
In the context of vector decomposition, one vector is often decomposed into two components: one component parallel to a given vector and another component orthogonal to it.
For example, if you have vector \(\textbf{v}\), decomposing it involves finding a vector \(\textbf{v}_{1}\) that is parallel to \(\textbf{w}\) and another vector \(\textbf{v}_{2}\) that is orthogonal to \(\textbf{w}\).
From the exercise, we determined \(\textbf{v}_{1} = \frac{5}{2} \textbf{w}\). To find the orthogonal component \(\textbf{v}_{2}\), we subtract \(\textbf{v}_{1}\) from \(\textbf{v}\): \(\textbf{v}_{2} = \textbf{v} - \textbf{v}_{1} = (2 \textbf{i} - 3 \textbf{j}) - (\frac{5}{2} \textbf{i} - \frac{5}{2} \textbf{j})\). Simplifying this gives: \(-\frac{1}{2} \textbf{i} - \frac{1}{2} \textbf{j}\).
Therefore, \(\textbf{v}_{2}\) is orthogonal to \(\textbf{w}\) and represents the part of \(\textbf{v}\) that does not align or project onto \(\textbf{w}\).
In the context of vector decomposition, one vector is often decomposed into two components: one component parallel to a given vector and another component orthogonal to it.
For example, if you have vector \(\textbf{v}\), decomposing it involves finding a vector \(\textbf{v}_{1}\) that is parallel to \(\textbf{w}\) and another vector \(\textbf{v}_{2}\) that is orthogonal to \(\textbf{w}\).
From the exercise, we determined \(\textbf{v}_{1} = \frac{5}{2} \textbf{w}\). To find the orthogonal component \(\textbf{v}_{2}\), we subtract \(\textbf{v}_{1}\) from \(\textbf{v}\): \(\textbf{v}_{2} = \textbf{v} - \textbf{v}_{1} = (2 \textbf{i} - 3 \textbf{j}) - (\frac{5}{2} \textbf{i} - \frac{5}{2} \textbf{j})\). Simplifying this gives: \(-\frac{1}{2} \textbf{i} - \frac{1}{2} \textbf{j}\).
Therefore, \(\textbf{v}_{2}\) is orthogonal to \(\textbf{w}\) and represents the part of \(\textbf{v}\) that does not align or project onto \(\textbf{w}\).
Component Form
Component form refers to expressing a vector in terms of its individual components. Typically, vectors are broken down into their \(\textbf{i}\) (x-axis) and \(\textbf{j}\) (y-axis) components.
This form is very useful in vector calculations because it allows for straightforward addition, subtraction, and projection operations.
Components of a vector \(\textbf{v}\) in 2D can be written as \(\textbf{v} = v_{x} \textbf{i} + v_{y} \textbf{j}\). Here, \(\textbf{i}\) and \(\textbf{j}\) are unit vectors along the x and y axes respectively, while \(\textbf{v}_{x}\) and \(\textbf{v}_{y}\) are the magnitudes of the vector in those directions.
In the given exercise, \(\textbf{v}\) and \(\textbf{w}\) are given in component form: \(\textbf{v} = 2 \textbf{i} - 3 \textbf{j}\) and \(\textbf{w} = \textbf{i} - \textbf{j}\). This format makes it easy to apply the projection formula and later the orthogonal decomposition.
Understanding component form is crucial because it allows us to see how vectors add together or subtract from one another, and how they project onto other vectors. By decomposing a vector into its components, we can handle more complex vector problems with simplicity and clarity.
This form is very useful in vector calculations because it allows for straightforward addition, subtraction, and projection operations.
Components of a vector \(\textbf{v}\) in 2D can be written as \(\textbf{v} = v_{x} \textbf{i} + v_{y} \textbf{j}\). Here, \(\textbf{i}\) and \(\textbf{j}\) are unit vectors along the x and y axes respectively, while \(\textbf{v}_{x}\) and \(\textbf{v}_{y}\) are the magnitudes of the vector in those directions.
In the given exercise, \(\textbf{v}\) and \(\textbf{w}\) are given in component form: \(\textbf{v} = 2 \textbf{i} - 3 \textbf{j}\) and \(\textbf{w} = \textbf{i} - \textbf{j}\). This format makes it easy to apply the projection formula and later the orthogonal decomposition.
Understanding component form is crucial because it allows us to see how vectors add together or subtract from one another, and how they project onto other vectors. By decomposing a vector into its components, we can handle more complex vector problems with simplicity and clarity.
Other exercises in this chapter
Problem 20
Plot each complex number in the complex plane and write it in polar form and in exponential form. $$ 9 \sqrt{3}+9 i $$
View solution Problem 20
Transform each polar equation to an equation in rectangular coordinates. Then identify and graph the equation. $$ r \cos \theta=4 $$
View solution Problem 21
Plot each complex number in the complex plane and write it in polar form and in exponential form. $$ 3-4 i $$
View solution Problem 21
Plot each point given in polar coordinates. $$ \left(3, \frac{\pi}{2}\right) $$
View solution