91Ó°ÊÓ

The dot product is an essential operation in vector calculus, allowing us to determine the relationship between two vectors. For two-dimensional vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), the dot product is calculated using the formula:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \)

In the original exercise, vectors \( \mathbf{a} = \langle 3, -1 \rangle \) and \( \mathbf{b} = \langle -4, 0 \rangle \) are considered. By substituting the values into the formula, we get:

• \( \mathbf{a} \cdot \mathbf{b} = 3 \times -4 + (-1) \times 0 = -12 \)

The dot product here tells us whether the vectors are perpendicular (if the dot product is zero), align in the same direction (if positive) or opposite directions (if negative). In this case, a negative result indicates that the vectors point in diverging directions.

Calculating the angle between vectors involves understanding the trigonometric relationship between the vectors. The angle \( \theta \) between any two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is determined by this fundamental relationship:

• \( \mathbf{a} \cdot \mathbf{b} = ||\mathbf{a}|| ||\mathbf{b}|| \cos(\theta) \)

From this equation, you can solve for \( \cos(\theta) \) to find:

• \( \cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{||\mathbf{a}|| ||\mathbf{b}||} \)

Applying these calculations to our exercise, we found that \( \cos(\theta) = \frac{-3}{\sqrt{10}} \). Knowing \( \cos(\theta) \) helps assess the nature of the angle itself.To actually find \( \theta \), we must use the inverse cosine function, which we will discuss next.

The magnitude of a vector, often represented as \( ||\mathbf{v}|| \), is a measure of its length, regardless of its direction. For a vector \( \mathbf{v} = \langle v_1, v_2 \rangle \), the magnitude is computed as:

• \( ||\mathbf{v}|| = \sqrt{v_1^2 + v_2^2} \)

This formula stems from the Pythagorean theorem. It provides a way to calculate the "size" of the vector in a two-dimensional space.For vector \( \mathbf{a} = \langle 3, -1 \rangle \), the magnitude is:

• \( ||\mathbf{a}|| = \sqrt{3^2 + (-1)^2} = \sqrt{10} \)

And for \( \mathbf{b} = \langle -4, 0 \rangle \), it is:

• \( ||\mathbf{b}|| = \sqrt{(-4)^2 + 0^2} = 4 \)

Knowing the magnitudes of vectors is critical for many operations, including finding the angle between them.

The inverse cosine function, denoted as \( \cos^{-1} \), allows us to determine an angle whose cosine is a given number. It is the mathematical tool used for finding an angle \( \theta \) between two vectors once you have calculated \( \cos(\theta) \).In the context of the exercise, we had calculated \( \cos(\theta) = \frac{-3}{\sqrt{10}} \). To find \( \theta \), we use:

• \( \theta = \cos^{-1}\left(\frac{-3}{\sqrt{10}}\right) \)

This operation yields an angle of approximately \( 1.89 \) radians. Understanding the inverse cosine is essential because it bridges the relationship between dot products and the angles between vectors. It helps conclude whether the angle is acute, with acute angles being less than \( \frac{\pi}{2} \) radians.