Problem 40 - Precalculus

Question

Find a unit vector in the direction of the given vector. $$v=(-10,24)$$

Step-by-Step Solution

Answer

The unit vector in the direction of \((-10, 24)\) is \(\left( \frac{-5}{13}, \frac{12}{13} \right)\).

Step 1: Understand the Problem

We need to find a unit vector in the direction of the given vector \( v = (-10, 24) \). A unit vector has a magnitude of 1 and points in the same direction as the vector we are given.

Step 2: Calculate the Magnitude of the Vector

To find the unit vector, we first need to calculate the magnitude of \( v \). The formula for the magnitude of a vector \( v = (x, y) \) is:\[ \|v\| = \sqrt{x^2 + y^2} \]Substituting the values, we get:\[ \|v\| = \sqrt{(-10)^2 + 24^2} = \sqrt{100 + 576} = \sqrt{676} = 26 \]

Step 3: Divide the Original Vector by its Magnitude

Now, divide each component of \( v \) by its magnitude to find the unit vector. The formula for finding a unit vector \( u \) in the direction of \( v \) is:\[ u = \frac{v}{\|v\|} = \left( \frac{-10}{26}, \frac{24}{26} \right) \]Simplifying each component:\[ u = \left( \frac{-5}{13}, \frac{12}{13} \right) \]

Step 4: Verify the Result

To verify, we can check that the magnitude of the unit vector \( u \) is 1:\[ \|u\| = \sqrt{\left( \frac{-5}{13} \right)^2 + \left( \frac{12}{13} \right)^2} = \sqrt{\frac{25}{169} + \frac{144}{169}} = \sqrt{\frac{169}{169}} = 1 \]This confirms our calculation is correct.

Key Concepts

Magnitude of a Vector

Calculating the magnitude of a vector is a crucial step when you're working with vectors, as it tells you how long, or how big a vector is.
This length is represented as a positive number. For a vector defined as \( v = (x, y) \), its magnitude \( \|v\| \) is computed using the formula:

• \[ \|v\| = \sqrt{x^2 + y^2} \]

This is essentially the Pythagorean theorem applied in a coordinate plane. To put it in simple terms, you're finding the hypotenuse of a right triangle where \( x \) and \( y \) are the lengths of the two legs.
In our example, for \( v = (-10, 24) \), the magnitude comes out to be 26. Recognizing the magnitude helps to convert vectors into a different form, such as unit vectors, which have a magnitude of 1.

Vector Components

A vector is composed of its components, which are essentially its directional values along each axis of a coordinate system.
These components are what define the vector's behavior and direction. For instance, in a 2D space, a vector \( v \) is described by its components \( v = (x, y) \).

• The first component \( x \) tells us the movement along the horizontal axis.
• The second component \( y \) signifies movement along the vertical axis.
This means each component plays a role in defining the vector's overall direction. In our example, the vector \( v = (-10, 24) \) is telling us it moves 10 units to the left and 24 units up. Knowing vector components is essential when performing operations like finding directional or unit vectors.

Directional Vectors

Directional vectors aim to provide a perspective on which way a vector is pointing.
When dealing with directional vectors, often what is desired is a unit vector in the same direction, mainly because it reduces the vector to its pure direction, eliminating size.Here's how you find a unit vector:

• First, compute the magnitude of the original vector.
• Next, divide each component of the original vector by this magnitude.
• The result will be a unit vector that maintains the direction of the original.

For example, if you are given a vector \( v = (-10, 24) \), we already calculated its magnitude as 26. To find the unit vector:

• Divide each component by the magnitude: \( u = \left( \frac{-10}{26}, \frac{24}{26} \right) \)

Simplified, the unit vector results in \( u = \left( \frac{-5}{13}, \frac{12}{13} \right) \).
This gives us a vector that has been resized to a magnitude of 1, while keeping the direction of the original vector.