Problem 42 - 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. \(\mathbf{v} = \langle 5, -12 \rangle\)

Step-by-Step Solution

Answer

The unit vector in the direction of the given vector \( \mathbf{v} = \langle 5, -12 \rangle \) is \( \mathbf{u} = \langle \frac{5}{13}, -\frac{12}{13} \rangle \), and its magnitude is 1.

Step 1: Calculating the Magnitude

First, calculate the magnitude or norm of the vector \( \mathbf{v} \). This can be achieved using the formula \( ||\mathbf{v}|| = \sqrt{v_1^2 + v_2^2} \), where \( v_1 \) and \( v_2 \) represent the components of the vector. Therefore, \( ||\mathbf{v}|| = \sqrt{5^2 + (-12)^2} = \sqrt{169} = 13.

Step 2: Find the Unit Vector

The unit vector \( \mathbf{u} \) in the direction of a given vector \( \mathbf{v} \) is calculated using the formula \( \mathbf{u} = \frac{\mathbf{v}}{||\mathbf{v}||} \). Divide each component of \( \mathbf{v} \) by its magnitude: \( \mathbf{u} = \frac{1}{13}\langle 5, -12 \rangle = \langle \frac{5}{13}, -\frac{12}{13} \rangle \).

Step 3: Verifying the Magnitude

Finally, verify that the result is indeed a unit vector by confirming that its magnitude equals 1. Compute \( ||\mathbf{u}|| = \sqrt{\left(\frac{5}{13}\right)^2 + \left(-\frac{12}{13}\right)^2} = 1 \).

Key Concepts

Vector Magnitude

When dealing with vectors, understanding their magnitude is fundamental. The magnitude of a vector, often referred to as its "length," is calculated using a basic formula. If you have a 2D vector like \( \mathbf{v} = \langle v_1, v_2 \rangle \), you find the magnitude by applying the Pythagorean theorem:

• Square each component of the vector.
• Add the squared values together.
• Take the square root of the resulting sum.

Formally, the magnitude \( ||\mathbf{v}|| \) is given by the equation \( \sqrt{v_1^2 + v_2^2} \). In our case with \( \mathbf{v} = \langle 5, -12 \rangle \), the calculation becomes \( \sqrt{5^2 + (-12)^2} \), which simplifies to \( \sqrt{169} = 13 \). The magnitude provides a sense of the vector's scale. Knowing how to find this is essential for comparing vectors and their sizes.

Direction of a Vector

Once we have the magnitude of a vector, we can begin to determine its direction. The direction of a vector is often represented by a unit vector. A unit vector is a vector that points in the same direction as the original vector but has a magnitude of 1.
To find a unit vector \( \mathbf{u} \), you divide each component of the vector by its magnitude. Mathematically, \( \mathbf{u} = \frac{\mathbf{v}}{||\mathbf{v}||} \). This operation does not change the direction but scales the vector to a length of exactly one unit.
For \( \mathbf{v} = \langle 5, -12 \rangle \), the unit vector is calculated as \( \mathbf{u} = \frac{1}{13}\langle 5, -12 \rangle = \langle \frac{5}{13}, -\frac{12}{13} \rangle \). This new vector \( \mathbf{u} \) maintains the direction of \( \mathbf{v} \), establishing a standard measure to describe direction irrespective of scale.

Magnitude Verification

After computing a unit vector, it’s important to verify that its magnitude equals 1. This step confirms that the unit vector correctly represents the direction of the original vector at a standardized length.
To verify this, calculate the magnitude of the unit vector \( \mathbf{u} \) using the same formula for vector magnitude. For \( \mathbf{u} = \langle \frac{5}{13}, -\frac{12}{13} \rangle \), we compute:

• Square each component: \( \left(\frac{5}{13}\right)^2 + \left(-\frac{12}{13}\right)^2 \).
• Sum the squared values.
• Take the square root of the sum, which will be \( 1 \).

If the magnitude is indeed 1, then the vector is confirmed to be a unit vector. This process ensures accuracy in representing the direction without altering the intrinsic qualities of the original vector.