Problem 33 - Advanced Engineering Mathematics

Question

In Problems \(33-36, \mathbf{a}=\langle 1,-1,3\rangle\) and \(\mathbf{b}=\langle 2,6,3\rangle .\) Find the indicated number. \(\operatorname{comp}_{\mathrm{b}}^{\mathbf{a}}\)

Step-by-Step Solution

Answer

The component of \( \mathbf{a} \) along \( \mathbf{b} \) is \( \frac{5}{7} \).

Step 1: Understand the Problem

The problem asks for the component of vector \( \mathbf{a} \) along vector \( \mathbf{b} \). This is known as the vector projection calculation, and involves finding a scalar component.

Step 2: Use the formula for the scalar component of a vector onto another

The formula to find the component of a vector \( \mathbf{a} \) along \( \mathbf{b} \) is given by: \[ \operatorname{comp}_{\mathbf{b}}^{\mathbf{a}} = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{b}\|} \]Here, \( \mathbf{a} \cdot \mathbf{b} \) is the dot product, and \( \|\mathbf{b}\| \) is the magnitude of vector \( \mathbf{b} \).

Step 3: Calculate the Dot Product \( \mathbf{a} \cdot \mathbf{b} \)

The dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated as:\[ \mathbf{a} \cdot \mathbf{b} = (1)(2) + (-1)(6) + (3)(3) \]Simplifying gives:\[ = 2 - 6 + 9 = 5 \]

Step 4: Calculate the Magnitude of Vector \( \mathbf{b} \)

The magnitude of vector \( \mathbf{b} \) is found using the formula:\[ \| \mathbf{b} \| = \sqrt{2^2 + 6^2 + 3^2} \]This simplifies to:\[ \| \mathbf{b} \| = \sqrt{4 + 36 + 9} = \sqrt{49} = 7 \]

Step 5: Compute the Scalar Component \( \operatorname{comp}_{\mathbf{b}}^{\mathbf{a}} \)

Using the values calculated:\[ \operatorname{comp}_{\mathbf{b}}^{\mathbf{a}} = \frac{5}{7} \]

Key Concepts

Understanding the Dot Product

The dot product is a fundamental operation in vector math. It involves multiplying two vectors to obtain a scalar. To calculate the dot product, you multiply corresponding components from two vectors and then add them together. Given vectors \( \mathbf{a} = \langle 1, -1, 3 \rangle \) and \( \mathbf{b} = \langle 2, 6, 3 \rangle \), the formula for the dot product is:

• \( \mathbf{a} \cdot \mathbf{b} = (1)(2) + (-1)(6) + (3)(3) \)
• Perform each multiplication: \( 1 \times 2 = 2 \), \( -1 \times 6 = -6 \), and \( 3 \times 3 = 9 \).
• Add these products together to get the final result: \( 2 - 6 + 9 = 5 \).

This scalar result, 5, tells us how much of \( \mathbf{a} \) is in the direction of \( \mathbf{b} \). The dot product is pivotal for further calculations such as finding projections or angles between vectors.

Magnitude of a Vector

The magnitude of a vector is essentially its length in space. For a vector \( \mathbf{b} = \langle 2, 6, 3 \rangle \), the magnitude can be calculated using the formula:

• \( \| \mathbf{b} \| = \sqrt{x^2 + y^2 + z^2} \).
• Plug in the components of \( \mathbf{b} \): \( x = 2, y = 6, z = 3 \).
• Calculate each squared component: \( 2^2 = 4 \), \( 6^2 = 36 \), \( 3^2 = 9 \).
• Sum these squares: \( 4 + 36 + 9 = 49 \).
• Take the square root to find the magnitude: \( \sqrt{49} = 7 \).

The magnitude, 7, gives a measure of vector \( \mathbf{b} \)'s length and is needed to compute the scalar component of one vector along another.

Scalar Component of a Vector

The scalar component, or projection of one vector onto another, tells us how much of one vector lies in the direction of another vector. To find this, we employ:

• Formula: \( \operatorname{comp}_{\mathbf{b}}^{\mathbf{a}} = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{b}\|} \).
• The dot product \( \mathbf{a} \cdot \mathbf{b} \) was calculated as 5.
• The magnitude \( \|\mathbf{b}\| \) was found to be 7.
• Divide the dot product by the magnitude: \( \frac{5}{7} \).

This gives a scalar value of \( \frac{5}{7} \), representing how much of \( \mathbf{a} \) aligns with \( \mathbf{b} \) in terms of length. This concept is especially useful in physics and engineering for analyzing forces and movements.