Problem 5
Question
True or False Given two nonzero, nonorthogonal vectors \(\mathbf{v}\) and \(\mathbf{w},\) it is always possible to decompose \(\mathbf{v}\) into two vectors, one parallel to \(\mathbf{w}\) and the other orthogonal to \(\mathbf{w}\).
Step-by-Step Solution
Verified Answer
True
1Step 1: Understand the decomposition requirement
To determine whether the statement is true or false, recall that any vector \(\mathbf{v}\) can be decomposed into two components: one that is parallel to \(\mathbf{w}\) and one that is orthogonal to \(\mathbf{w}\).
2Step 2: Find the parallel component
The parallel component of \(\mathbf{v}\) with respect to \(\mathbf{w}\) is given by: \[ \mathbf{v}_{\parallel} = \frac{\mathbf{v} \cdot \mathbf{w} }{\mathbf{w} \cdot \mathbf{w}} \mathbf{w} \] where \ \cdot \ denotes the dot product.
3Step 3: Find the orthogonal component
The orthogonal component of \(\mathbf{v}\) with respect to \(\mathbf{w}\) can be found by subtracting the parallel component from \(\mathbf{v}\): \[ \mathbf{v}_{\perp} = \mathbf{v} - \mathbf{v}_{\parallel} \]
4Step 4: Verify orthogonality
To ensure \(\mathbf{v}_{\perp}\) is orthogonal to \(\mathbf{w}\), compute the dot product: \[ \mathbf{v}_{\perp} \cdot \mathbf{w} = (\mathbf{v} - \mathbf{v}_{\parallel}) \cdot \mathbf{w} \]\ \ \ Since \(\mathbf{v}_{\parallel}\) is parallel to \(\mathbf{w}\), \ \mathbf{v}_{\parallel} \cdot \mathbf{w} = \mathbf{w} \cdot \mathbf{v}_{\parallel} \ will hold. Therefore, \(\mathbf{v}_{\perp}\) is indeed orthogonal to \(\mathbf{w}\).
5Step 5: Conclude the problem
Given that our steps show such a decomposition (parallel and orthogonal) is always possible, the original statement is true.
Key Concepts
Parallel ComponentOrthogonal ComponentDot Product
Parallel Component
The parallel component of a vector \(\mathbf{v}\) with respect to another vector \(\mathbf{w}\) is a crucial concept in vector decomposition. This component represents the part of \(\mathbf{v}\) that lies in the same direction as \(\mathbf{w}\). To find the parallel component of \(\mathbf{v}\), we use the formula:
\[ \mathbf{v}_{\parallel} = \frac{\mathbf{v} \cdot \mathbf{w} }{\mathbf{w} \cdot \mathbf{w}} \mathbf{w} \]
Here, the term \( \mathbf{v} \cdot \mathbf{w} \) represents the dot product of vectors \(\mathbf{v}\) and \(\mathbf{w}\). This formula helps project \(\mathbf{v}\) onto \(\mathbf{w}\).
When you break down the formula:
- \( \mathbf{v} \cdot \mathbf{w} \) measures how much of \(\mathbf{v}\) points in the direction of \(\mathbf{w}\).
- \( \mathbf{w} \cdot \mathbf{w} \) is essentially the magnitude of \(\mathbf{w}\) squared and normalizes the projection.
So, \(\mathbf{v}_{\parallel}\) gives you a new vector that shows the directional influence of \(\mathbf{w}\) on \(\mathbf{v}\). This tells us how \(\mathbf{v}\) behaves along \(\mathbf{w}\)'s path.
\[ \mathbf{v}_{\parallel} = \frac{\mathbf{v} \cdot \mathbf{w} }{\mathbf{w} \cdot \mathbf{w}} \mathbf{w} \]
Here, the term \( \mathbf{v} \cdot \mathbf{w} \) represents the dot product of vectors \(\mathbf{v}\) and \(\mathbf{w}\). This formula helps project \(\mathbf{v}\) onto \(\mathbf{w}\).
When you break down the formula:
- \( \mathbf{v} \cdot \mathbf{w} \) measures how much of \(\mathbf{v}\) points in the direction of \(\mathbf{w}\).
- \( \mathbf{w} \cdot \mathbf{w} \) is essentially the magnitude of \(\mathbf{w}\) squared and normalizes the projection.
So, \(\mathbf{v}_{\parallel}\) gives you a new vector that shows the directional influence of \(\mathbf{w}\) on \(\mathbf{v}\). This tells us how \(\mathbf{v}\) behaves along \(\mathbf{w}\)'s path.
Orthogonal Component
The orthogonal component of a vector \(\mathbf{v}\) with respect to \(\mathbf{w}\) tells us how \(\mathbf{v}\) deviates perpendicularly from \(\mathbf{w}\). This component is found by subtracting the parallel component from the original vector:
\[ \mathbf{v}_{\perp} = \mathbf{v} - \mathbf{v}_{\parallel} \]
Computing this difference leaves us with the part of \(\mathbf{v}\) that does not align with \(\mathbf{w}\). Additionally, to verify that the orthogonal component is truly orthogonal to \(\mathbf{w}\), we check the dot product:
\[ \mathbf{v}_{\perp} \cdot \mathbf{w} = (\mathbf{v} - \mathbf{v}_{\parallel}) \cdot \mathbf{w} \]
If this dot product equals \(0\), then \(\mathbf{v}_{\perp}\) is indeed orthogonal to \(\mathbf{w}\). Since \(\mathbf{v}_{\parallel}\) is parallel to \(\mathbf{w}\), \(\mathbf{v}_{\parallel} \cdot \mathbf{w}\) will hold true and ensure orthogonality. The orthogonal component provides a perpendicular snapshot of how \(\mathbf{v}\) exists outside the influence of \(\mathbf{w}\).
\[ \mathbf{v}_{\perp} = \mathbf{v} - \mathbf{v}_{\parallel} \]
Computing this difference leaves us with the part of \(\mathbf{v}\) that does not align with \(\mathbf{w}\). Additionally, to verify that the orthogonal component is truly orthogonal to \(\mathbf{w}\), we check the dot product:
\[ \mathbf{v}_{\perp} \cdot \mathbf{w} = (\mathbf{v} - \mathbf{v}_{\parallel}) \cdot \mathbf{w} \]
If this dot product equals \(0\), then \(\mathbf{v}_{\perp}\) is indeed orthogonal to \(\mathbf{w}\). Since \(\mathbf{v}_{\parallel}\) is parallel to \(\mathbf{w}\), \(\mathbf{v}_{\parallel} \cdot \mathbf{w}\) will hold true and ensure orthogonality. The orthogonal component provides a perpendicular snapshot of how \(\mathbf{v}\) exists outside the influence of \(\mathbf{w}\).
Dot Product
The dot product is a fundamental operation for understanding vector relationships. It takes two vectors and returns a scalar. Mathematically, it is defined as:
\[ \mathbf{v} \cdot \mathbf{w} = v_{1}w_{1} + v_{2}w_{2} + ... + v_{n}w_{n} \]
Here’s what the dot product can tell us:
- **Magnitude Influence:** How much one vector influences another in terms of direction and length.
- **Angle Information:** If the dot product is zero, it means the vectors are orthogonal, as their directionality shares no common path.
For vectors \(\mathbf{v}\) and \(\mathbf{w}\), the dot product is used to compute both the parallel and orthogonal components:
- **Parallel Component:** \( \mathbf{v}_{\parallel} = \frac{\mathbf{v} \cdot \mathbf{w} }{\mathbf{w} \cdot \mathbf{w}} \mathbf{w} \) ensures alignment projection.
- **Verifying Orthogonality:** Checking \(\mathbf{v}_{\perp} \cdot \mathbf{w} = 0\) confirms perpendicularity.
The dot product essentially serves as a bridge, linking vector magnitudes and directional influences through scalar multiplication.
\[ \mathbf{v} \cdot \mathbf{w} = v_{1}w_{1} + v_{2}w_{2} + ... + v_{n}w_{n} \]
Here’s what the dot product can tell us:
- **Magnitude Influence:** How much one vector influences another in terms of direction and length.
- **Angle Information:** If the dot product is zero, it means the vectors are orthogonal, as their directionality shares no common path.
For vectors \(\mathbf{v}\) and \(\mathbf{w}\), the dot product is used to compute both the parallel and orthogonal components:
- **Parallel Component:** \( \mathbf{v}_{\parallel} = \frac{\mathbf{v} \cdot \mathbf{w} }{\mathbf{w} \cdot \mathbf{w}} \mathbf{w} \) ensures alignment projection.
- **Verifying Orthogonality:** Checking \(\mathbf{v}_{\perp} \cdot \mathbf{w} = 0\) confirms perpendicularity.
The dot product essentially serves as a bridge, linking vector magnitudes and directional influences through scalar multiplication.
Other exercises in this chapter
Problem 4
Is the sine function even, odd, or neither?
View solution Problem 4
Draw the angle \(\frac{5 \pi}{6}\) in standard position.
View solution Problem 5
In the complex plane, the \(x\) -axis is referred to as the _________ axis, and the \(y\) -axis is called the _______ axis.
View solution Problem 5
\(\sin \frac{5 \pi}{4}=\) ______
View solution