Problem 36 - Precalculus : Building Concepts and Connections

Question

Find a unit vector in the same direction as the given vector. $$\mathbf{v}=\langle-12,5\rangle$$

Step-by-Step Solution

Answer

The unit vector in the same direction as \( \mathbf{v}=\langle-12,5\rangle \) is \( \mathbf{u} = \langle \frac{-12}{13}, \frac{5}{13} \rangle \).

Step 1: Find the magnitude of the vector

The first step is to find the magnitude (or length) of the vector \( \mathbf{v} \). The magnitude of a vector \( \mathbf{v} = \langle x, y \rangle \) is calculated as \( \sqrt{x^2 + y^2} \). So for the vector \( \mathbf{v} = \langle -12, 5 \rangle \), its magnitude is \( \sqrt{(-12)^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \).

Step 2: Divide the vector by its magnitude

Next, divide each component of the vector by the magnitude. This will give a new vector \( \mathbf{u} = \langle \frac{x}{\text{magnitude}}, \frac{y}{\text{magnitude}} \rangle \). So for our vector, \( \mathbf{u} = \langle \frac{-12}{13}, \frac{5}{13} \rangle \).

Step 3: Check the magnitude of the unit vector

To verify the result, we can check if the magnitude of the unit vector \( \mathbf{u} \) is 1. The magnitude of \( \mathbf{u} = \langle \frac{-12}{13}, \frac{5}{13} \rangle \) is \( \sqrt{(-\frac{12}{13})^2 + (\frac{5}{13})^2} = 1 \). So, \( \mathbf{u} \) is indeed a unit vector.

Key Concepts

Vector Magnitude

To start with the concept of vector magnitude, it's integral to understand that vectors are fundamental entities in fields such as physics, engineering, and computer graphics. They possess not only direction, but also a quantity known as magnitude, which is a measure of a vector's length. In our exercise, we examined the vector
\( \mathbf{v} = \langle -12, 5 \rangle \), and to find its magnitude, we employed the Pythagorean theorem tailored for two-dimensional vectors:
\[ \sqrt{x^2 + y^2} \]. Specifically, for \( \mathbf{v} \), this results in:
\[ \sqrt{(-12)^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \].
The magnitude gives us an indication of how 'long' the vector is, which is essential when determining how to scale the vector to a unit vector.

Vector Normalization

Moving onward to vector normalization, this is a process by which a vector is converted to a unit vector—one that maintains its direction but has a magnitude of 1. This is a foundational step in many applications that require a pure direction without an associated scale. To normalize a vector, we divide each of its components by the vector's magnitude.
Following our example, we normalized vector \( \mathbf{v} \) by dividing its components by 13, its magnitude:
\[ \mathbf{u} = \langle \frac{-12}{13}, \frac{5}{13} \rangle \].
By performing this operation, we simplify the components of the vector to their minimum expression while retaining the original direction, a crucial step in the normalization process. Remember, it's always vital to ensure the magnitude of the normalized vector equates to 1 to confirm correct normalization.

Vector Components

Lastly, let's discuss vector components, which are the building blocks of a vector. Each component represents the vector's influence along a particular axis in its coordinate system. In two dimensions, a vector's components correspond to changes along the x-axis and y-axis.
In the context of our problem, the vector \( \mathbf{v} = \langle -12, 5 \rangle \) has a component of -12 in the x-direction and 5 in the y-direction. These values tell us how far the vector moves in each axis the space it's defined in. Breaking down vectors into components is not just useful for visualization, but it's also essential when performing operations such as addition, subtraction, or when projecting vectors onto axes.