Problem 9 - Precalculus Mathematics for Calculus

Question

Find \((a) u \cdot v\) and \((b)\) the angle between \(u\) and \(v\) to the nearest degree. $$\mathbf{u}=\langle 3,-2\rangle, \quad \mathbf{v}=\langle 1,2\rangle$$

Step-by-Step Solution

Answer

(a) \(-1\); (b) \(\theta \approx 99\) degrees.

Step 1: Find the Dot Product

To find the dot product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \), we use the formula for dot product: \( u \cdot v = u_1 \times v_1 + u_2 \times v_2 \). For these vectors \( \mathbf{u} = \langle 3, -2 \rangle \) and \( \mathbf{v} = \langle 1, 2 \rangle \), calculate:\[ u \cdot v = 3 \times 1 + (-2) \times 2 = 3 - 4 = -1 \].Thus, the dot product \( u \cdot v = -1 \).

Step 2: Calculate Magnitude of Vectors

The magnitude of a vector \( \mathbf{u} = \langle u_1, u_2 \rangle \) is given by \( ||\mathbf{u}|| = \sqrt{u_1^2 + u_2^2} \). Calculate the magnitudes of \( \mathbf{u} \) and \( \mathbf{v} \): \[ ||\mathbf{u}|| = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \] \[ ||\mathbf{v}|| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \].

Step 3: Find the Cosine of the Angle

The cosine of the angle \( \theta \) between two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is given by the formula: \[ \cos(\theta) = \frac{u \cdot v}{||\mathbf{u}|| \times ||\mathbf{v}||} \].Substitute the values calculated:\[ \cos(\theta) = \frac{-1}{\sqrt{13} \times \sqrt{5}} = \frac{-1}{\sqrt{65}} \].

Step 4: Calculate the Angle

To find \( \theta \), take the arccos (inverse cosine) of the value from Step 3:\[ \theta = \cos^{-1} \left( \frac{-1}{\sqrt{65}} \right) \].Use a calculator to evaluate \( \theta \) and convert it to degrees. The angle \( \theta \approx 99 \) degrees.

Key Concepts

Vector Magnitude

Understanding vector magnitude is essential when working with vector mathematics. The magnitude of a vector quantifies its length, or how far it "stretches" in space. To visualize, think of a vector as an arrow pointing from one point to another.
The magnitude is then the length of that arrow. The formula to calculate the magnitude of a vector \( \mathbf{u} = \langle u_1, u_2 \rangle \) in two dimensions is \( ||\mathbf{u}|| = \sqrt{u_1^2 + u_2^2} \). This formula is derived from the Pythagorean theorem, considering the vector as a hypotenuse of a right triangle formed by its components.

• For example, the vector \( \mathbf{u} = \langle 3, -2 \rangle \) would have a magnitude of \( \sqrt{3^2 + (-2)^2} = \sqrt{13} \).
• Similarly, the vector \( \mathbf{v} = \langle 1, 2 \rangle \) would have a magnitude of \( \sqrt{1^2 + 2^2} = \sqrt{5} \).

This concept helps us to not only understand how "long" a vector is but also to perform further operations such as finding the angle between two vectors.

Cosine of Angle Between Vectors

Once we know the dot product and the magnitudes of the vectors, we can find the cosine of the angle between them. This is a crucial step in determining the actual angle between two vectors.The cosine of the angle \( \theta \) between vectors \( \mathbf{u} \) and \( \mathbf{v} \) is determined using the formula: \[ \cos(\theta) = \frac{u \cdot v}{||\mathbf{u}|| \times ||\mathbf{v}||} \]Where:

• \( u \cdot v \) is the dot product of the vectors,
• \( ||\mathbf{u}|| \) and \( ||\mathbf{v}|| \) are their magnitudes.

For vectors \( \mathbf{u} = \langle 3, -2 \rangle \) and \( \mathbf{v} = \langle 1, 2 \rangle \), the dot product is \(-1\), and their magnitudes were found to be \( \sqrt{13} \) and \( \sqrt{5} \) respectively. Substituting these into our formula gives:\[ \cos(\theta) = \frac{-1}{\sqrt{13} \times \sqrt{5}} = \frac{-1}{\sqrt{65}} \]This result shows the relationship between the vectors' directions. A value closer to 1 or -1 indicates that the vectors are nearly parallel, either in the same or opposite directions.

Arccos Function

The arccosine function helps us find the angle corresponding to a given cosine value. It's the inverse operation of the cosine and is vital in trigonometry when working with vectors.Given a cosine value, the arccos function \( \cos^{-1} \) returns the angle whose cosine is the specified value. For example, if \( \cos(\theta) \) is known, the angle \( \theta \) can be computed by:\[ \theta = \cos^{-1}(\cos(\theta)) \]In the context of vectors \( \mathbf{u} \) and \( \mathbf{v} \), after finding \( \cos(\theta) = \frac{-1}{\sqrt{65}} \), we determine \( \theta \) by:\[ \theta = \cos^{-1} \left( \frac{-1}{\sqrt{65}} \right) \]Use a calculator to find the angle in degrees. In this example, \( \theta \approx 99 \text{ degrees} \). This step helps translate the mathematical cosine value into a comprehensible angle in space, showing the degree of separation between \( \mathbf{u} \) and \( \mathbf{v} \).