Problem 35 - Precalculus

Question

Find a unit vector in the direction of the given vector. $$\mathbf{v}=(-5,-12)$$

Step-by-Step Solution

Answer

The unit vector is \( \left( \frac{-5}{13}, \frac{-12}{13} \right) \).

Step 1: Calculate the Magnitude of the Given Vector

To find a unit vector in the direction of the given vector \( \mathbf{v} = (-5, -12) \), we first need to calculate the magnitude of \( \mathbf{v} \). The magnitude (or length) of a vector \( \mathbf{v} = (x, y) \) is calculated by the formula:\[||\mathbf{v}|| = \sqrt{x^2 + y^2}\]Substitute the given values:\[||\mathbf{v}|| = \sqrt{(-5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13\]

Step 2: Normalize the Vector

To find the unit vector, we divide each component of the vector \( \mathbf{v} \) by its magnitude. The formula for the unit vector \( \mathbf{u} \) in the direction of \( \mathbf{v} \) is given by:\[\mathbf{u} = \left( \frac{x}{||\mathbf{v}||}, \frac{y}{||\mathbf{v}||} \right)\]Substituting \( x = -5 \), \( y = -12 \), and \( ||\mathbf{v}|| = 13 \):\[\mathbf{u} = \left( \frac{-5}{13}, \frac{-12}{13} \right)\]

Step 3: Provide the Unit Vector

The unit vector in the direction of \( \mathbf{v} \) is:\[\mathbf{u} = \left( \frac{-5}{13}, \frac{-12}{13} \right)\]

Key Concepts

Vector Magnitude

The magnitude of a vector, often denoted as \( ||\mathbf{v}|| \), is a measure of its length. Imagine a vector as an arrow drawn in a plane; the magnitude tells us how long that arrow is. When we calculate the magnitude of a vector, we are essentially finding the distance from the origin point \((0, 0)\) to the point represented by the vector \((x, y)\). This is done using a straightforward formula derived from the Pythagorean theorem:

• For a vector \( \mathbf{v} = (x, y) \), its magnitude is given by \( ||\mathbf{v}|| = \sqrt{x^2 + y^2} \).
• In the example \( \mathbf{v} = (-5, -12) \), the magnitude is \( ||\mathbf{v}|| = \sqrt{(-5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \).

This computation is crucial because the magnitude serves as a scaling factor to convert any vector to a unit vector, which has a magnitude of 1.

Normalize Vector

To normalize a vector means to convert it into a unit vector. A unit vector is simply a vector with a magnitude of 1 that points in the same direction as the original vector. The reason we normalize vectors is to keep the direction intact while simplifying the magnitude to unity.

• For any vector \( \mathbf{v} = (x, y) \), the unit vector \( \mathbf{u} \) is calculated by dividing each component of \( \mathbf{v} \) by its magnitude: \( \mathbf{u} = \left( \frac{x}{||\mathbf{v}||}, \frac{y}{||\mathbf{v}||} \right) \).
• Taking the vector \( (-5, -12) \) with magnitude 13, the unit vector becomes \( \mathbf{u} = \left( \frac{-5}{13}, \frac{-12}{13} \right) \).

By normalizing, we attained a vector that points exactly in the same original direction but is easier to work with when analyzing direction or performing further calculations.

Direction of a Vector

The direction of a vector is an essential aspect when we talk about vectors in physics and mathematics. It shows where the vector points with respect to a chosen reference point, typically the origin. The direction is maintained even when we scale the vector by a constant, like when we normalize it.

• Relative to vectors and coordinates, direction can often be inferred or represented using angles, but in simple vector component form, it suffices to talk about the direction using unit vectors.
• When using a unit vector, like \( \mathbf{u} = \left( \frac{-5}{13}, \frac{-12}{13} \right) \), the components directly define the direction in which the vector points.

A unit vector directly communicates direction because its length is 1. This means it fully carries the directional information of the original vector \( \mathbf{v} \) without complicating the scale.