Problem 17 - College Trigonometry

Question

In Exercises \(1-20\), use the pair of vectors \(\vec{v}\) and \(\vec{w}\) to find the following quantities. $$ \vec{v}=\left\langle\frac{1}{2}, \frac{\sqrt{3}}{2}\right\rangle \text { and } \vec{w}=\left\langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle $$

Step-by-Step Solution

Answer

Dot product: \(\frac{-\sqrt{2} + \sqrt{6}}{4}\); Magnitudes: both 1.

Step 1: Understanding Vectors and Their Components

Here, we need to work with two vectors: \(\vec{v} = \left\langle\frac{1}{2}, \frac{\sqrt{3}}{2}\right\rangle\) and \(\vec{w} = \left\langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle\). Each vector is represented by its components in a two-dimensional space. The components of \(\vec{v}\) are \(v_1 = \frac{1}{2}\) and \(v_2 = \frac{\sqrt{3}}{2}\), while for \(\vec{w}\), the components are \(w_1 = -\frac{\sqrt{2}}{2}\) and \(w_2 = \frac{\sqrt{2}}{2}\). Our task is to compute specific quantities involving these vectors.

Step 2: Calculating the Dot Product

The dot product of two vectors \(\vec{v} = \langle a_1, a_2 \rangle\) and \(\vec{w} = \langle b_1, b_2 \rangle\) is calculated using the formula \(a_1b_1 + a_2b_2\). For our vectors, this becomes \(\frac{1}{2} \times \left(-\frac{\sqrt{2}}{2}\right) + \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}\). Simplifying, we calculate the dot product: \(-\frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4}\). Therefore, the dot product \(\vec{v} \cdot \vec{w} = \frac{-\sqrt{2} + \sqrt{6}}{4}\).

Step 3: Calculating the Magnitude of Vectors

To find the magnitude of a vector \(\vec{v} = \langle a_1, a_2 \rangle\), use the formula \(\sqrt{a_1^2 + a_2^2}\). For \(\vec{v}\), this becomes \(\sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}\), which simplifies to \(\sqrt{\frac{1}{4} + \frac{3}{4}}\). Thus, the magnitude \(||\vec{v}|| = 1\). Similarly, for \(\vec{w}\), the magnitude is \(\sqrt{\left(-\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2}\), which simplifies to \(\sqrt{\frac{2}{4} + \frac{2}{4}} = \sqrt{1}\). Therefore, \(||\vec{w}|| = 1\).

Step 4: Angle Between the Vectors Using Dot Product

The angle \(\theta\) between two vectors can be calculated using the formula \(\cos \theta = \frac{\vec{v} \cdot \vec{w}}{||\vec{v}|| \cdot ||\vec{w}||}\). We've calculated \(\vec{v} \cdot \vec{w} = \frac{-\sqrt{2} + \sqrt{6}}{4}\) in Step 2, and both magnitudes are 1. Therefore, \(\cos \theta = \frac{\frac{-\sqrt{2} + \sqrt{6}}{4}}{1 \cdot 1} = \frac{-\sqrt{2} + \sqrt{6}}{4}\). To find \(\theta\), take \(\cos^{-1}\) of that result.

Key Concepts

Dot Product

The dot product is a fundamental operation that gives us the measure of how long one vector stretches in the direction of another. When working with vectors, such as \( \vec{v} = \langle \frac{1}{2}, \frac{\sqrt{3}}{2} \rangle \) and \( \vec{w} = \langle -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \rangle \), calculating the dot product helps us understand their directional relationship.

To calculate the dot product, multiply corresponding components of the vectors and add the resulting products:

• First, calculate: \( \frac{1}{2} \times \left(-\frac{\sqrt{2}}{2}\right) = -\frac{\sqrt{2}}{4} \)
• Then, calculate: \( \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2} = \frac{\sqrt{6}}{4} \)

Add these results to get the dot product: \(-\frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} = \frac{-\sqrt{2} + \sqrt{6}}{4} \).

This scalar value gives us insight into the angle between the vectors and their magnitudes. It is also a crucial step in finding the angle between the vectors, which we will explore next.

Magnitude of Vectors

The magnitude of a vector, also known as length or norm, measures how long the vector is. This is crucial in understanding the size of the vector without regard to its direction. Let's calculate the magnitude for vectors \( \vec{v} \) and \( \vec{w} \).

For \( \vec{v} = \langle \frac{1}{2}, \frac{\sqrt{3}}{2} \rangle \):

• First, square each component: \( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \) and \( \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \)
• Add the squares: \( \frac{1}{4} + \frac{3}{4} = 1 \)

Take the square root: \( \sqrt{1} = 1 \). Hence, the magnitude of \( \vec{v} \) is 1.

Similarly, for \( \vec{w} = \langle -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \rangle \):

• Square each component: \( \left(-\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} \) and \( \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} \)
• Add the squares: \( \frac{2}{4} + \frac{2}{4} = 1 \)

Again, take the square root: \( \sqrt{1} = 1 \). The magnitude of \( \vec{w} \) is also 1.

Understanding magnitude helps in visualizing vectors' size, which is important when comparing vectors or when using them in further calculations involving angles.

Angle Between Vectors

Finding the angle \theta\u0019 between two vectors allows us to understand their directional divergence. We use the dot product and magnitudes to find this angle, employing the formula:

\[ \cos \theta = \frac{\vec{v} \cdot \vec{w}}{||\vec{v}|| \cdot ||\vec{w}||} \]
For our vectors, we calculated the dot product \( \vec{v} \cdot \vec{w} = \frac{-\sqrt{2} + \sqrt{6}}{4} \) and found both magnitudes \( ||\vec{v}|| \) and \( ||\vec{w}|| \) to be 1.

Substitute these into the angle formula:

• \( \cos \theta = \frac{\frac{-\sqrt{2} + \sqrt{6}}{4}}{1 \cdot 1} = \frac{-\sqrt{2} + \sqrt{6}}{4} \)

To find \( \theta \), compute the arccosine (inverse cosine) of this value.

The resultant angle tells us how far apart the directions of the vectors are from each other, helping in applications such as physics, engineering, and computer graphics, where directional relationships are key.