Problem 23 - Algebra and Trigonometry

Question

23-28 (a) Calculate proj, \(\mathbf{u} .\) (b) Resolve \(\mathbf{u}\) into \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\) where \(\mathbf{u}_{1}\) is parallel to \(\mathbf{v}\) and \(\mathbf{u}_{2}\) is orthogonal to \(\mathbf{v}\). $$\mathbf{u}=\langle- 2,4\rangle, \quad \mathbf{v}=\langle 1,1\rangle$$

Step-by-Step Solution

Answer

(a) \(\text{proj}_{\mathbf{v}} \, \mathbf{u} = \langle 1, 1 \rangle\); (b) \(\mathbf{u}_1 = \langle 1, 1 \rangle\), \(\mathbf{u}_2 = \langle -3, 3 \rangle\).

Step 1: Calculate the Projection Formula

The projection of vector \(\mathbf{u}\) onto vector \(\mathbf{v}\) is given by the formula: \[ \text{proj}_{\mathbf{v}} \, \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \cdot \mathbf{v} \]First, we'll compute the dot products needed.

Step 2: Compute the Dot Products

Calculate the dot product \(\mathbf{u} \cdot \mathbf{v}\):\[(-2) \cdot 1 + 4 \cdot 1 = -2 + 4 = 2\]Calculate the dot product \(\mathbf{v} \cdot \mathbf{v}\):\[1 \cdot 1 + 1 \cdot 1 = 1 + 1 = 2\]

Step 3: Calculate the Projection

Use the dot products to find \(\text{proj}_{\mathbf{v}} \, \mathbf{u}\):\[\text{proj}_{\mathbf{v}} \, \mathbf{u} = \frac{2}{2} \cdot \langle 1, 1 \rangle = 1 \cdot \langle 1, 1 \rangle = \langle 1, 1 \rangle\]Thus, the projection of \(\mathbf{u}\) onto \(\mathbf{v}\) is \(\langle 1, 1 \rangle\).

Step 4: Resolve \(\mathbf{u}\) into \(\mathbf{u}_1\) and \(\mathbf{u}_2\)

Since \(\mathbf{u}_1\) is parallel to \(\mathbf{v}\), we have \(\mathbf{u}_1 = \text{proj}_{\mathbf{v}} \, \mathbf{u} = \langle 1, 1 \rangle\).To find \(\mathbf{u}_2\), use:\[\mathbf{u}_2 = \mathbf{u} - \mathbf{u}_1 = \langle -2, 4 \rangle - \langle 1, 1 \rangle = \langle -3, 3 \rangle\]Thus, \(\mathbf{u}\) is resolved into \(\mathbf{u}_1 = \langle 1, 1 \rangle\) and \(\mathbf{u}_2 = \langle -3, 3 \rangle\).

Key Concepts

Dot Product

The dot product is a fundamental operation used in vector mathematics. It combines two vectors to yield a single scalar value.
This scalar indicates how "aligned" the vectors are. In our problem, the dot product \( \mathbf{u} \cdot \mathbf{v} \) is calculated as follows:

• Multiply the corresponding components of the vectors: \( (-2) \times 1 \) and \( 4 \times 1 \).
• Add the resultant values: \(-2 + 4 = 2\).

Understanding the dot product is crucial as it helps determine both the projection of one vector onto another and their orthogonality.
If the dot product is zero, the vectors are orthogonal (perpendicular). However, in our scenario, the result is 2, indicating the vectors are not orthogonal.

Vector Resolution

Vector resolution involves breaking down a vector into its component parts based on a given direction. This can be visualized as dividing a vector \( \mathbf{u} \) into two parts: one parallel and one perpendicular to another vector \( \mathbf{v} \).
In our example:

• \( \mathbf{u}_1 \), the parallel component, is found using the projection formula:
  \( \text{proj}_{\mathbf{v}} \, \mathbf{u} = \langle 1, 1 \rangle \).
• The orthogonal component, \( \mathbf{u}_2 \), is discovered by subtracting \( \mathbf{u}_1 \) from \( \mathbf{u} \):
  \( \mathbf{u}_2 = \mathbf{u} - \mathbf{u}_1 = \langle -3, 3 \rangle \).

This process allows us to fully break down \( \mathbf{u} \) into its distinct directional parts, making it easier to analyze its behavior relative to \( \mathbf{v} \).

Orthogonal Vectors

Orthogonal vectors are vectors that are perpendicular to each other. This means their dot product equals zero.
In vector analysis, recognizing orthogonal vectors is vital as they have significant applications in fields like physics and engineering.
For the given vectors, \( \mathbf{u} \) and \( \mathbf{v} \), we resolved \( \mathbf{u} \) into:

• \( \mathbf{u}_1 = \langle 1, 1 \rangle \)
• \( \mathbf{u}_2 = \langle -3, 3 \rangle \)

While \( \mathbf{u}_1 \) aligns with \( \mathbf{v} \), the vector \( \mathbf{u}_2 \) is orthogonal to \( \mathbf{v} \).
This distinction helps in dissecting \( \mathbf{u} \) into its fundamental components, showcasing its dual nature of alignment and perpendicularity with respect to \( \mathbf{v} \). Orthogonality often assists in simplifying complex vector operations and understanding spatial relationships.