Problem 39 - Precalculus with Limits

Question

In Exercises 39-48, find a unit vector in the direction of the given vector. Verify that the result has a magnitude of 1. \(\mathbf{u} = \langle 3, 0 \rangle\)

Step-by-Step Solution

Answer

The unit vector in the direction of \( \mathbf{u} = \langle 3, 0 \rangle \) is \( \langle 1, 0 \rangle \).

Step 1: Calculating the magnitude of the original vector

The magnitude (sometimes also called 'length' or 'norm') of a vector \( \mathbf{v} = \langle a, b \rangle \) can be found using the formula \( ||\mathbf{v}|| = \sqrt{a^2 + b^2} \). Here, applying this formula to the vector \( \mathbf{u} = \langle 3, 0 \rangle \), we find \( ||\mathbf{u}|| = \sqrt{3^2 + 0^2} = \sqrt{9} = 3 \).

Step 2: Finding the unit vector

The unit vector in the same direction as a given vector is found by dividing the original vector by its magnitude. Therefore, the unit vector \( \mathbf{u_{unit}} \) in the direction of \( \mathbf{u} \) is given by \( \mathbf{u_{unit}} = \mathbf{u} / ||\mathbf{u}|| = \langle 3/3, 0/3 \rangle = \langle 1, 0 \rangle \).

Step 3: Verifying the results

To ensure the obtained vector is a unit vector, its magnitude should be equal to 1. Calculating for the vector \( \langle 1, 0 \rangle \), it gives \( ||\mathbf{u_{unit}}|| = \sqrt{1^2 + 0^2} = \sqrt{1} = 1 \). Thus, the magnitude of the calculated unit vector is 1, verifying that the solution is correct.

Key Concepts

magnitude of a vector

Vectors are fundamental in mathematics and physics, representing quantities that have both magnitude and direction. The magnitude of a vector, often denoted as \(||\mathbf{v}||\), represents the "length" or "size" of the vector. This is a crucial step when dealing with vectors since it helps quantify their scale.
To find the magnitude of a vector in a two-dimensional plane, such as \( \mathbf{v} = \langle a, b \rangle \), we use the formula:

• \(||\mathbf{v}|| = \sqrt{a^2 + b^2}\)

This formula comes from the Pythagorean theorem, reflecting how a vector can be seen as the hypotenuse of a right triangle with sides of length \(a\) and \(b\).
Take, for instance, the vector \( \mathbf{u} = \langle 3, 0 \rangle \). To find its magnitude, substitute the values into the formula:

• \(||\mathbf{u}|| = \sqrt{3^2 + 0^2} = \sqrt{9} = 3\)

This calculation shows that the magnitude of \(\mathbf{u}\) is 3, indicating the length of the vector from the origin (0,0) to the point (3,0) on the coordinate plane.

vector normalization

To convert a vector into a unit vector, we normalize it. A unit vector is a vector that has a magnitude of 1 but points in the same direction as the original vector. This process is called vector normalization and is especially useful in various applications of science and engineering.
Normalization involves scaling the original vector so that its length is 1. Here's how you do it:

• Take the original vector \( \mathbf{v} = \langle a, b \rangle \)
• Divide each of its components by the magnitude of the vector \(||\mathbf{v}||\).

So, the unit vector \( \mathbf{v_{unit}} \) is given by:

• \( \mathbf{v_{unit}} = \left\langle \frac{a}{||\mathbf{v}||}, \frac{b}{||\mathbf{v}||} \right\rangle \)

For example, with vector \( \mathbf{u} = \langle 3, 0 \rangle \) and magnitude \||\mathbf{u}|| = 3\, the unit vector \( \mathbf{u_{unit}} \) becomes:

• \(\mathbf{u_{unit}} = \left\langle \frac{3}{3}, \frac{0}{3} \right\rangle = \langle 1, 0 \rangle\)

This process ensures the vector maintains its original direction but has a set length, which is crucial for simplifying vector operations in calculations.

direction of a vector

The direction of a vector is an integral part of its identity, indicating where and how the vector points in the space. While vectors have both magnitude and direction, the direction alone is determined by the "angle" the vector makes with a reference direction, usually the positive x-axis.
When considering direction, particularly in unit vectors, it is the slope or orientation of the vector rather than its length that we are interested in. A unit vector keeps only the directional information, stripped of scale, making it particularly useful in cases where you only want to describe where something is heading, not how far.
To find the direction of a vector \( \mathbf{v} = \langle a, b \rangle \), you can use trigonometry. For instance, by determining the angle \( \theta \) it forms with the x-axis using:

• \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \)

For our example vector \( \mathbf{u} = \langle 3, 0 \rangle \), since \(b = 0\), it lies along the x-axis, indicating that the directional angle \( \theta \) is 0.
Direction is key in vector operations, ensuring vectors point the correct way in solving physics problems, computer graphics, and navigation challenges.