Problem 17 - Precalculus

Question

Find the angle (round to the nearest degree) between each pair of vectors. $$\langle-4,6\rangle \text { and }\langle-6,8\rangle$$

Step-by-Step Solution

Answer

The angle is approximately 3 degrees.

Step 1: Compute the Dot Product

\(\mathbf{u} \cdot \mathbf{v} = (-4)(-6) + (6)(8) = 24 + 48 = 72\)

Step 2: Compute the Magnitudes

\(|\mathbf{u}| = \sqrt{(-4)^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}\)
\(|\mathbf{v}| = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\)

Step 3: Find the Angle

\(\cos\theta = \frac{72}{2\sqrt{13} \cdot 10} = \frac{72}{20\sqrt{13}} = \frac{3.6}{\sqrt{13}} \approx \frac{3.6}{3.6056} \approx 0.9985\)
\(\theta = \cos^{-1}(0.9985) \approx 3^\circ\)

Key Concepts

Dot Product

Understanding the dot product is essential when calculating the angle between two vectors. The dot product, also known as the scalar product, is a way of multiplying two vectors to produce a single number, or scalar. This value can give insight into the relationship between the vectors. For instance, if the dot product is positive, the vectors point roughly in the same direction, negative means they point in opposite directions, and zero indicates they are perpendicular.

To find the dot product of vectors \( \mathbf{a} = \langle -4, 6 \rangle \) and \( \mathbf{b} = \langle -6, 8 \rangle \), use this formula:

• Find the product of their corresponding components: \( -4 \times -6 \) and \( 6 \times 8 \).
• Add these products together: \( (-4 \times -6) + (6 \times 8) \).

Using the values, compute: \( 24 + 48 = 72 \). So, the dot product of the vectors is \( 72 \). This value is pivotal for finding the angle between the two vectors.

Vectors in Two Dimensions

Vectors in two dimensions are essential in representing physical quantities requiring both direction and magnitude, like force or velocity. In this context, a vector is often denoted by two numbers in a pair of parentheses or angle brackets, representing its components along the x-axis and y-axis.

For instance, the vector \( \langle -4, 6 \rangle \) describes a movement of 4 units left and 6 units up in a Cartesian plane. Similarly, \( \langle -6, 8 \rangle \) moves 6 units left and 8 units up.

When working with vectors, understanding their geometric representation is crucial. Each vector can be visualized as an arrow pointing from the origin to a specific point. The direction of the arrow shows the vector's direction, and the length of the arrow reflects the vector's magnitude.

• The magnitude of a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \) is calculated using the formula: \( \sqrt{a_1^2 + a_2^2} \).
• Applying it to our vectors: \( \text{Magnitude of } \langle -4, 6 \rangle = \sqrt{(-4)^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} \).
• Similarly, \( \text{Magnitude of } \langle -6, 8 \rangle = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \).

Understanding these properties and calculations prepares us for using them in further computations, such as finding angles.

Trigonometry

In the context of vectors, trigonometry helps us find angles between them using the dot product and magnitudes. The formula that relates these elements is:
\[ \cos \theta = \frac{ \mathbf{a} \cdot \mathbf{b} }{ \| \mathbf{a} \| \| \mathbf{b} \| } \]where:

• \( \mathbf{a} \cdot \mathbf{b} \) is the dot product of the vectors.
• \( \| \mathbf{a} \| \) and \( \| \mathbf{b} \| \) are the magnitudes of the vectors.
• \( \theta \) is the angle between the vectors.

Using our previous calculations:- Dot product: \( 72 \)- Magnitude of \( \mathbf{a} \): \( \sqrt{52} \)- Magnitude of \( \mathbf{b} \): \( 10 \)

We find:\[ \cos \theta = \frac{72}{\sqrt{52} \times 10} = \frac{72}{10\sqrt{52}} \]Now, calculate \( \theta \) by taking the inverse cosine, usually denoted as \( \cos^{-1} \), of the result:\( \theta \approx \cos^{-1}\left( \frac{72}{10\sqrt{52}} \right) \).

After solving, round the angle to the nearest degree to obtain the final answer. Understanding these steps allows you to effectively use trigonometry in vector operations.