Problem 9 - Essential Calculus Early Transcendentals

Question

2-10 Find \(\mathbf{a} \cdot \mathbf{b}\) $$|\mathbf{a}|=6, \quad|\mathbf{b}|=5, \quad$$ the angle between a and \(\mathbf{b}\) is 2\(\pi / 3\)

Step-by-Step Solution

Answer

The dot product \(\mathbf{a} \cdot \mathbf{b}\) is \(-15\).

Step 1: Recall the Dot Product Formula

The dot product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) can be calculated using the formula: \(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(\theta)\) where \(\theta\) is the angle between the vectors.

Step 2: Plug in the Magnitudes and Angle

Given that \(|\mathbf{a}| = 6\), \(|\mathbf{b}| = 5\), and the angle \(\theta = \frac{2\pi}{3}\), substitute these values into the formula: \[\mathbf{a} \cdot \mathbf{b} = 6 \times 5 \times \cos\left(\frac{2\pi}{3}\right)\].

Step 3: Calculate the Cosine of the Angle

Calculate \(\cos\left(\frac{2\pi}{3}\right)\). The cosine of \(\frac{2\pi}{3}\) is \(-\frac{1}{2}\).

Step 4: Compute the Dot Product

With \(\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\), calculate the dot product: \[\mathbf{a} \cdot \mathbf{b} = 6 \times 5 \times \left(-\frac{1}{2}\right) = -15\].

Key Concepts

Understanding Vector Magnitude

In vector mathematics, the term *magnitude* refers to the length or size of a vector. Imagine a vector as an arrow pointing in space; the magnitude is simply how long this arrow is. It gives us a numeric representation of the vector's size, usually shown as \(|\mathbf{a}|\).
\[\text{For example, for a vector } \mathbf{a} = (x, y, z), |\mathbf{a}| = \sqrt{x^2 + y^2 + z^2}.\]
This formula is derived from the Pythagorean theorem and applies to three-dimensional vectors. For two-dimensional vectors, \(|\mathbf{a}| = \sqrt{x^2 + y^2}\).

• **Importance**: Understanding the magnitude helps in assessing how significant a vector is in terms of size.
• **Physical Significance**: In physics, magnitudes can represent quantities like force, velocity, or displacement.

In our problem, the given magnitudes \(|\mathbf{a}| = 6\) and \(|\mathbf{b}| = 5\) are applied directly into the dot product formula.

Exploring the Angle Between Vectors

The angle between two vectors is critical in understanding the relationship between them. It tells us how "aligned" or "intersecting" two vectors are. In a geometric sense, imagine each vector as arrows stemming from a common origin. The angle is the measure of rotation needed to overlap one vector with the other.
To find this angle, we use the dot product formula:
\[\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(\theta)\]
where \(\theta\) is the angle between the vectors. In practical terms, if \(\theta = 0\degree\), the vectors are perfectly aligned, resulting in a maximum dot product. A \(90\degree\) angle suggests vectors are orthogonal, meaning no direct relation in directionality.

• This angle helps in fields like physics and computer graphics to determine properties like vector alignment and projection.
• Knowing \(\theta\) allows one to resolve vectors into components and understand orientations.

In this exercise, \(\theta = \frac{2\pi}{3}\) indicates that the vectors are neither aligned nor orthogonal, influencing the sign and magnitude of the dot product.

The Role of Cosine in Determining Vector Relationships

The cosine of the angle \(\theta\) between two vectors is a central aspect of understanding their interaction. Cosine values range from -1 to 1, showing how vectors align:

• A cosine of 1 means vectors are in the same direction.
• A cosine of 0 suggests they are orthogonal.
• A negative cosine, like \(-\frac{1}{2}\) in this case, implies partial opposition.

Using \(\cos(\frac{2\pi}{3}) = -\frac{1}{2}\), it signifies that vectors \(\mathbf{a}\) and \(\mathbf{b}\) are pushing against each other to some extent, indicated by the negative result of the dot product calculation. This cosine value scales the dot product, reflecting both the angle's effect and the product of the vector magnitudes. Understanding this relationship helps in predicting vector behavior, particularly in fields like engineering and physics.