Problem 25 - 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 \(\mathbf{v}\). $$\mathbf{u}=\langle 1,2\rangle, \quad \mathbf{v}=\langle 1,-3\rangle$$

Step-by-Step Solution

Answer

(a) \( \text{proj}_{\mathbf{v}} \mathbf{u} = \langle -0.5, 1.5 \rangle \), (b) \( \mathbf{u}_1 = \langle -0.5, 1.5 \rangle \), \( \mathbf{u}_2 = \langle 1.5, 0.5 \rangle \).

Step 1: Calculate the Dot Product

To calculate the projection of \( \mathbf{u} \) onto \( \mathbf{v} \), we first need the dot product of \( \mathbf{u} \) and \( \mathbf{v} \). Use the formula for the dot product: \[ \mathbf{u} \cdot \mathbf{v} = (1)(1) + (2)(-3) = 1 - 6 = -5 \]

Step 2: Find the Magnitude of v Squared

Next, find the magnitude of \( \mathbf{v} \) squared, which is needed for the projection formula:\[ ||\mathbf{v}||^2 = (1)^2 + (-3)^2 = 1 + 9 = 10 \]

Step 3: Calculate the Projection of u onto v

Use the projection formula for \( \text{proj}_{\mathbf{v}} \mathbf{u} \):\[ \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{v}||^2} \times \mathbf{v} = \frac{-5}{10} \times \langle 1, -3 \rangle = \langle -0.5, 1.5 \rangle \]

Step 4: Resolve u into Two Components

\( \mathbf{u}_1 \) is the component of \( \mathbf{u} \) parallel to \( \mathbf{v} \), which is just \( \text{proj}_{\mathbf{v}} \mathbf{u} \). We have:\( \mathbf{u}_1 = \langle -0.5, 1.5 \rangle \).The orthogonal component \( \mathbf{u}_2 \) is given by:\[ \mathbf{u}_2 = \mathbf{u} - \mathbf{u}_1 = \langle 1, 2 \rangle - \langle -0.5, 1.5 \rangle = \langle 1 + 0.5, 2 - 1.5 \rangle = \langle 1.5, 0.5 \rangle \]

Step 5: Solution Summary

We found that \( \text{proj}_{\mathbf{v}} \mathbf{u} = \langle -0.5, 1.5 \rangle \). \( \mathbf{u} \) is resolved into \( \mathbf{u}_1 = \langle -0.5, 1.5 \rangle \) and \( \mathbf{u}_2 = \langle 1.5, 0.5 \rangle \), with \( \mathbf{u}_1 \) parallel to \( \mathbf{v} \) and \( \mathbf{u}_2 \) orthogonal to \( \mathbf{v} \).

Key Concepts

Dot Product

The dot product is a fundamental operation when working with vectors in mathematics and physics. It is the product of two vectors, providing a scalar value that indicates the degree to which the vectors align.
To compute the dot product of vectors \( \mathbf{u} = \langle a, b \rangle \) and \( \mathbf{v} = \langle c, d \rangle \), use the formula:

• \( \mathbf{u} \cdot \mathbf{v} = ac + bd \)

For example, with \( \mathbf{u} = \langle 1, 2 \rangle \) and \( \mathbf{v} = \langle 1, -3 \rangle \), the dot product is \( 1 \times 1 + 2 \times (-3) = 1 - 6 = -5 \).
This value helps determine projections, find angles between vectors, and assess vector parallelism. A dot product of zero indicates orthogonal vectors, meaning they meet at a right angle.

Magnitude of a Vector

Understanding the magnitude of a vector helps us measure its length or size. The magnitude of vector \( \mathbf{v} = \langle c, d \rangle \) is calculated using the Pythagorean theorem, expressed as:

• \( ||\mathbf{v}|| = \sqrt{c^2 + d^2} \)

To find the magnitude of \( \mathbf{v} = \langle 1, -3 \rangle \), compute \( ||\mathbf{v}|| = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \).
This magnitude is crucial when projecting one vector onto another, as it appears in the denominator of the projection formula. Calculating projection involves not just determining alignment but understanding the component sizes relative to the other vector.

Orthogonal Vectors

Orthogonal vectors are vectors that meet at a 90-degree angle, which implies having no influence on one another. This means their dot product is zero. Mathematically, two vectors \( \mathbf{a} \) and \( \mathbf{b} \) are orthogonal if \( \mathbf{a} \cdot \mathbf{b} = 0 \).
When resolving a vector \( \mathbf{u} \) into two components \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \), where \( \mathbf{u}_2 \) is orthogonal to another vector \( \mathbf{v} \), you subtract the component of \( \mathbf{u} \) that is parallel to \( \mathbf{v} \).
This gives the orthogonal component. For example, \( \mathbf{u}_2 = \mathbf{u} - \mathbf{u}_1 \).

• In our case, \( \mathbf{u}_2 = \langle 1, 2 \rangle - \langle -0.5, 1.5 \rangle = \langle 1.5, 0.5 \rangle \)

This representation helps in various applications like graphics, where orthogonal directions help define independent dimensional axes.

Parallel Vectors

Parallel vectors have the same or opposite direction. They do not need to have the same magnitude, but their directional ratios remain constant. Vectors \( \mathbf{a} \) and \( \mathbf{b} \) are parallel if there exists a scalar \( k \) such that \( \mathbf{a} = k \mathbf{b} \).
When resolving vectors, the component parallel to another vector embodies the shared direction scaled by the projection.
For instance, the parallel component \( \mathbf{u}_1 \) of vector \( \mathbf{u} \) with relation to \( \mathbf{v} \) is found using:

• \( \mathbf{u}_1 = \text{proj}_\mathbf{v} \mathbf{u} \)
• In this case, \( \mathbf{u}_1 = \langle -0.5, 1.5 \rangle \)

Using the projection, one can analyze how much of \( \mathbf{u} \) points in the direction of \( \mathbf{v} \), offering insights in physics and engineering, such as force or velocity components in specific directions.