Problem 49 - Precalculus with Limits

Question

Using the Angle Between Two Vectors Exercises \(49-52,\) find \(u \cdot v,\) where \(\theta\) is the angle between \(\mathbf{u}\) and \(v .\) $$\|\mathbf{u}\|=4,\|\mathbf{v}\|=10, \theta=\frac{2 \pi}{3}$$

Step-by-Step Solution

Answer

-20

Step 1: Write down the given information

First, we identify the given magnitudes of the vectors and the angle between them. These are: \(\|\mathbf{u}\|=4, \|\mathbf{v}\|=10, \theta=\frac{2 \pi}{3}\)

Step 2: Identify the formula for calculating the dot product

We want to find \(u \cdot v\) (Dot product of vectors \(\mathbf{u}\) and \(\mathbf{v}\)). By definition, the dot product of two vectors is given by the product of their magnitudes and the cosine of the angle between them, which can be written formally like this: \(u \cdot v = \|\mathbf{u}\|* \|\mathbf{v}\| * \cos(\theta)\).

Step 3: Substitute the given values into the formula

Now that we have the formula, we can substitute our values into the formula to find the dot product: \(u \cdot v = 4 * 10 * \cos\left(\frac{2 \pi}{3}\right) = 40 \cos\left(\frac{2 \pi}{3}\right)\)

Step 4: Evaluate the cosine function and compute product

We need to evaluate the \(\cos\left(\frac{2 \pi}{3}\right)\). This value is -0.5. So we multiply out all values (40 * -0.5 = -20).

Key Concepts

Vector Magnitude

In the world of mathematics and physics, the magnitude of a vector is essentially a measure of its length. For a vector \textbf{u} represented by coordinates in two-dimensional space, its magnitude can be calculated using the formula: \( \| \mathbf{u} \| = \sqrt{u_x^2 + u_y^2} \), where \( u_x \) and \( u_y \) are the vector's horizontal and vertical components, respectively.

Think of it as the distance from the starting point (usually the origin) to the point defined by the vector. Even in three-dimensional space, the concept remains the same but includes the additional z-component: \( \| \mathbf{u} \| = \sqrt{u_x^2 + u_y^2 + u_z^2} \).

To put this in the context of our exercise, for vector \( \mathbf{u} \) with a magnitude of 4, we can imagine a four units long arrow drawn from the origin point.

Understanding vector magnitude is critical because it's a fundamental component in various vector operations, including the computation of the dot product, which combines magnitudes with the angle between vectors to yield a scalar value.

Angle Between Vectors

The angle between vectors is a key concept in understanding their orientation and interaction. This concept is not about how long the vectors are (their magnitudes), but rather how they're positioned with respect to one another. To visualize this, imagine two arrows originating from the same point, and the space between these arrows is what we refer to as the angle between the vectors.

The angle is important in physics and engineering; for example, it can represent the difference in direction between forces or velocities.

In our problem, the angle between vectors \( \mathbf{u} \) and \( \mathbf{v} \) is \( \frac{2 \pi}{3} \), which is 120 degrees. Understanding the angle is essential for calculating the dot product as it tells us how directional forces may interact to propel an object along its path, which directly ties into our next concept, the cosine function.

Cosine Function

The cosine function, a pivotal concept in trigonometry, connects the angle between two vectors with their dot product. It is a way to determine the horizontal component of a vector when it is rotated at a certain angle from the horizontal axis.

In the context of the dot product, the cosine function takes the angle between the two vectors and provides a scalar value that scales the product of their magnitudes. This scalar can be positive, negative, or zero. A positive value means the vectors are oriented less than 90 degrees from each other, a negative value indicates more than 90 degrees, and a value of zero signifies perpendicular vectors.

The dot product formula \( u \cdot v = \|\mathbf{u}\|* \|\mathbf{v}\| * \cos(\theta) \) relies on the cosine of the angle to determine how much one vector extends in the direction of the other. In our exercise, the negative value of the cosine \( \cos\left(\frac{2 \pi}{3}\right) = -0.5 \) suggests that vectors \( \mathbf{u} \) and \( \mathbf{v} \) point in more divergent directions, which translates to their dot product being a negative number, reflecting a sort of 'opposition' in their directions.