Problem 25 - Precalculus with Limits

Question

In Exercises 25-30, use the dot product to find the magnitude of \(\mathbf{u}\). \(\mathbf{u} = \langle -8, 15 \rangle\)

Step-by-Step Solution

Answer

The magnitude of vector \(\mathbf{u} = \langle -8, 15 \rangle\) is \(||\mathbf{u}|| = \sqrt{(-8)^2 + (15)^2}\)

Step 1: Understand the given vector

The given vector \(\mathbf{u} = \langle -8, 15 \rangle\), contains two components: -8, 15.

Step 2: Apply the dot product formula

The dot product of a vector with itself is the square of its magnitude. Therefore, we can compute the magnitude of a vector by getting the square root of the dot product of the vector with itself. The dot product of \(\mathbf{u} = \langle -8, 15 \rangle\) with itself is calculated as: \((\mathbf{u}\cdot\mathbf{u}) = (-8)^2 + (15)^2\).

Step 3: Calculate the magnitude

After calculating the dot product, the magnitude of vector \(\mathbf{u}\) is then the square root of the result obtained: \(||\mathbf{u}|| = \sqrt{(-8)^2 + (15)^2}\)

Key Concepts

Magnitude of a Vector

The magnitude of a vector is a measure of its length. It tells us how long the vector is, regardless of its direction. To find the magnitude of a vector, we can use the dot product. When a vector is dotted with itself, it equals the square of its magnitude. For example, the vector \(\mathbf{u} = \langle -8, 15 \rangle\) can be used to find its magnitude by applying the formula:

• First, compute the dot product with itself: \(\mathbf{u} \cdot \mathbf{u} = (-8)^2 + (15)^2\)
• Next, find the square root of this result: \(||\mathbf{u}|| = \sqrt{(-8)^2 + (15)^2}\)

Using this approach helps simplify the calculation, providing the clear distance of a vector from the origin in a coordinate system.

Vector Components

Vectors have components that describe their position in space relative to the axes. For a two-dimensional vector like \(\mathbf{u} = \langle -8, 15 \rangle\), these components are \(-8\) on the x-axis and \(15\) on the y-axis. Each component contributes to the overall direction and length of the vector. By understanding components:

• The horizontal component, in this case, is \(-8\). It tells us how far and in which direction the vector moves along the x-axis.
• The vertical component is \(15\), indicating movement along the y-axis.

When working with vector problems, breaking them down into components makes analysis easier, especially in complex scenarios with multiple vectors. Components come into play when applying various operations such as addition, scalar multiplication, and especially when calculating the dot product.

Precalculus Exercises

Precalculus exercises involving vectors prepare students for more advanced mathematical concepts. Understanding how to calculate the magnitude, work with vector components, and use the dot product are foundational skills. In exercises like the one involving the vector \(\mathbf{u} = \langle -8, 15 \rangle\), students practice:

• Identifying components of vectors.
• Using the dot product to find the magnitude.
• Developing skills that are essential in calculus, physics, and engineering.

These exercises are crucial as they build a solid understanding of vectors, which are vectors are extensively used in real-life applications such as navigation, computer graphics, and mechanics. Mastery of the precalculus fundamentals ensures success in tackling more complex mathematical and scientific problems.