Problem 42 - Precalculus with Limits

Question

Finding a Unit Vector In Exercises \(39-48,\) find a unit vector in the direction of the given vector. Verify that the result has a magnitude of 1 . $$\mathbf{v}=\langle 5,-12\rangle$$

Step-by-Step Solution

Answer

The unit vector in the direction of the vector \(\mathbf{v}=\langle 5,-12\rangle\) is \(\mathbf{u}=\langle \frac{5}{13},\frac{-12}{13} \rangle\). The magnitude of this unit vector has been verified to be 1.

Step 1: Calculating the magnitude of the given vector

First, the magnitude (or length) of vector \(\mathbf{v}\) should be calculated. The magnitude of a vector \(\mathbf{v}=\langle x,y\rangle\) in 2D space is computed as \(\|\mathbf{v}\| =\sqrt{x^2 + y^2}\). In this case, \(x=5\) and \(y=-12\). Therefore, the magnitude \(\|\mathbf{v}\| =\sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169}=13\)

Step 2: Deriving the unit vector

Once the magnitude is found, the unit vector in the direction of \(\mathbf{v}\) can be calculated by dividing each component of the vector by its magnitude. Hence, the unit vector is \(\mathbf{u}=\langle \frac{x}{\|\mathbf{v}\|},\frac{y}{\|\mathbf{v}\|} \rangle = \langle \frac{5}{13},\frac{-12}{13} \rangle\)

Step 3: Verifying the magnitude of the unit vector

To ensure that the result is a unit vector, the magnitude of \(\mathbf{u}\) must be checked. The magnitude \(\|\mathbf{u}\| = \sqrt{ (\frac{5}{13})^2 + (\frac{-12}{13})^2 } = \sqrt{ (\frac{25}{169}) + (\frac{144}{169}) } = \sqrt{1}=1\). Because the magnitude of the unit vector is 1, the given result is correct.

Key Concepts

Magnitude of a Vector

Understanding the magnitude of a vector is crucial for various fields such as physics, engineering, and computer graphics. If we picture a vector as an arrow, the magnitude represents the length of this arrow, which signifies how far the vector reaches in a certain direction.

The formula to calculate the magnitude of a two-dimensional vector \(\mathbf{v} = \langle x, y \rangle\) is given by \(\|\mathbf{v}\| = \sqrt{x^2 + y^2}\), where \(x\) and \(y\) are the vector's horizontal and vertical components, respectively. Just like in the provided exercise, when you're given a vector \(\mathbf{v} = \langle 5, -12 \rangle\), the magnitude is the square root of the sum of the squares of its components: \(\|\mathbf{v}\| = \sqrt{5^2 + (-12)^2}\). It turns out to be \(\sqrt{169} = 13\).

This value is essential because it describes the 'size' or 'reach' of the vector, irrespective of its direction.

Vector Components

Going further into detail, each vector can be broken down into its vector components which illustrate how much the vector extends along each axis in a given coordinate system. For instance, in two dimensions, a vector \(\mathbf{v}\) has two components, \(\mathbf{v} = \langle x, y \rangle\).

These components can be visualized by drawing a right triangle with the vector as the hypotenuse. The horizontal leg represents the \(x\)-component, and the vertical leg represents the \(y\)-component. Thus, in our exercise example \(\mathbf{v} = \langle 5, -12 \rangle\), \(5\) is the \(x\)-component, showing movement along the x-axis, and \(\-12\) is the \(y\)-component, indicating movement along the y-axis but in the opposite direction due to the negative sign.

Understanding vector components is fundamental when performing operations such as vector addition, multiplication, or finding a unit vector, as they are the basis from which these calculations are made.

Normalizing a Vector

Normalizing, also known as creating a unit vector, is the process of rescaling a vector to have a magnitude of 1 without changing its direction. It's a common procedure in vector calculus and computer science, especially in graphics where direction needs to be maintained but length may interfere with computations.

To normalize a vector \(\mathbf{v}\), you simply divide each of its components by the vector's magnitude. Following our earlier example, the unit vector \(\mathbf{u}\) from \(\mathbf{v} = \langle 5, -12 \rangle\) is found by dividing each component by its magnitude, \(13\): \(\mathbf{u} = \langle \frac{5}{13}, \frac{-12}{13} \rangle\).

Finally, you verify the unit vector by checking if it has a magnitude of 1, which it does in this case: \(\|\mathbf{u}\| = \sqrt{(\frac{5}{13})^2 + (\frac{-12}{13})^2} = 1\). This normalization process enables us to simplify the vector's representation while preserving its essential direction characteristics, facilitating calculations that concern direction over magnitude.