Problem 33 - Algebra and Trigonometry

Question

(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 \(\mathbf{v} .\) $$ \mathbf{u}=\langle 2,9\rangle, \quad \mathbf{v}=\langle- 3,4\rangle $$

Step-by-Step Solution

Answer

The projection is \( \langle -\frac{18}{5}, \frac{24}{5} \rangle \), and \( \mathbf{u} \) resolves into \( \mathbf{u}_1 = \langle -\frac{18}{5}, \frac{24}{5} \rangle \), \( \mathbf{u}_2 = \langle \frac{28}{5}, \frac{21}{5} \rangle \).

Step 1: Calculate Dot Product

Calculate the dot product of \( \mathbf{u} \) and \( \mathbf{v} \), which is \( \mathbf{u} \cdot \mathbf{v} = 2(-3) + 9(4) = -6 + 36 = 30 \).

Step 2: Calculate Magnitude Squared of \( \mathbf{v} \)

Find \( \|\mathbf{v}\|^2 \), the magnitude squared of \( \mathbf{v} \). This is \( (-3)^2 + 4^2 = 9 + 16 = 25 \).

Step 3: Calculate Projection

Use the formula for the projection of \( \mathbf{u} \) onto \( \mathbf{v} \) given by \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} \mathbf{v} \). Substitute the values: \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{30}{25} \langle -3, 4 \rangle = \frac{6}{5} \langle -3, 4 \rangle = \langle -\frac{18}{5}, \frac{24}{5} \rangle \).

Step 4: Resolve \( \mathbf{u} \) into Components

The component of \( \mathbf{u} \) that is parallel to \( \mathbf{v} \) is \( \mathbf{u}_1 = \langle -\frac{18}{5}, \frac{24}{5} \rangle \). The component that is orthogonal is \( \mathbf{u}_2 = \mathbf{u} - \mathbf{u}_1 = \langle 2, 9 \rangle - \langle -\frac{18}{5}, \frac{24}{5} \rangle = \langle 2 + \frac{18}{5}, 9 - \frac{24}{5} \rangle = \langle \frac{28}{5}, \frac{21}{5} \rangle \).

Key Concepts

Dot Product

The dot product is a mathematical operation that takes two equal-length sequences of numbers and returns a single number. In the context of vectors, like \( \mathbf{u} = \langle 2, 9 \rangle \) and \( \mathbf{v} = \langle -3, 4 \rangle \), the dot product is determined by multiplying the corresponding components of each vector and then summing the results.
Calculating \( \mathbf{u} \cdot \mathbf{v} \) involves:

• Multiplying the x-components: \( 2 \times -3 = -6 \)
• Multiplying the y-components: \( 9 \times 4 = 36 \)
• Adding these results: \( -6 + 36 = 30 \)

The resulting dot product of 30 not only helps in further computations, like finding projections, but it also gives insight into the angle between the vectors.
If the dot product is positive, the angle is acute, meaning the vectors are pointing in a roughly similar direction. If it is zero, the vectors are orthogonal (at right angles). Finally, if it’s negative, it means vectors are pointing in more opposite directions.

Magnitude of a Vector

The magnitude of a vector gives the length or size of the vector, oftentimes represented as \( \| \mathbf{v} \| \). For a vector with components \( \langle x, y \rangle \), the magnitude is found using the Pythagorean theorem: \( \| \mathbf{v} \| = \sqrt{x^2 + y^2} \).
In our example, the magnitude squared of vector \( \mathbf{v} = \langle -3, 4 \rangle \) is calculated as follows:

• Square each component: \( (-3)^2 = 9 \) and \( 4^2 = 16 \)
• Add these squares: \( 9 + 16 = 25 \)

This value, 25, is actually the magnitude squared of the vector, not the magnitude itself. However, note that when projecting vectors, the magnitude squared is used to simplify division steps.
Understanding the magnitude of a vector is crucial as it provides the scalar quantity of distance irrespective of direction.

Orthogonal Components

Orthogonal components refer to vectors that are perpendicular to each other, indicating they do not share any common directionality. When we resolve a vector, like \( \mathbf{u} = \langle 2, 9 \rangle \), into components relative to another vector \( \mathbf{v} = \langle -3, 4 \rangle \), the orthogonal component \( \mathbf{u}_2 \) is the part of \( \mathbf{u} \) that remains when the parallel component is subtracted.
To find this, we take:

• The vector \( \mathbf{u} \)
• Subtract its parallel projection onto \( \mathbf{v} \) which is \( \mathbf{u}_1 \)
• Resulting in \( \mathbf{u}_2 \)

In simplified terms, \( \mathbf{u}_2 = \mathbf{u} - \mathbf{u}_1 \), where \( \mathbf{u}_1 \) is the component of \( \mathbf{u} \) that is along the same direction as \( \mathbf{v} \), and \( \mathbf{u}_2 \) is the residue orthogonal to it.
Understanding this concept is important since orthogonal vectors have practical applications such as in physics, where they describe forces acting independently.

Vector Resolution

Vector resolution is the process of breaking down a vector into multiple component vectors. The original vector is expressed as a sum of these components, often making complex vector calculations simpler and more intuitive.
When resolving \( \mathbf{u} = \langle 2, 9 \rangle \) relative to \( \mathbf{v} = \langle -3, 4 \rangle \), we identify \( \mathbf{u}_1 \), the part parallel to \( \mathbf{v} \), and \( \mathbf{u}_2 \), the orthogonal part.
The steps include:

• Calculate \( \mathbf{u}_1 \) as \( \text{proj}_{\mathbf{v}} \mathbf{u} \).
• Then determine \( \mathbf{u}_2 \) by subtracting \( \mathbf{u}_1 \) from \( \mathbf{u} \).

This breakdown helps in practical calculations, like those in physics and engineering, where components along certain axes are required.
Vector resolution allows us to work with simpler parts, giving clear insights into the nature of forces, velocities, or other vector quantities in different directions.