Problem 43 - Contemporary Precalculus

Question

Find a unit vector that has the same direction as \(v\). $$\langle 4,-5\rangle$$

Step-by-Step Solution

Answer

Question: Find the unit vector in the same direction as the given vector \(v = \langle 4, -5 \rangle\). Answer: \(\langle \frac{4}{\sqrt{41}}, \frac{-5}{\sqrt{41}} \rangle\)

Step 1: Find the magnitude of the given vector

Calculate the magnitude of vector \(v = \langle 4,-5 \rangle \) using the formula: $$\|v\| = \sqrt{4^2 + (-5)^2}$$

Step 2: Calculate the magnitude

By evaluating the expression within the square root, we get: $$\|v\| = \sqrt{16 + 25} = \sqrt{41}$$

Step 3: Determine the unit vector

Now, we will divide each component of \(v\) by its magnitude \(\|v\|\) to get the unit vector in the same direction: $$u = \frac{1}{\|v\|}v = \frac{1}{\sqrt{41}}\langle 4, -5 \rangle = \langle \frac{4}{\sqrt{41}}, \frac{-5}{\sqrt{41}} \rangle$$ The unit vector in the same direction as \(v\) is \(\langle \frac{4}{\sqrt{41}}, \frac{-5}{\sqrt{41}} \rangle\).

Key Concepts

Vector Magnitude

The concept of vector magnitude refers to the length or size of a vector. It's like finding out how long a line segment is in geometry. The magnitude of a vector is usually denoted as \( \|v\| \) and can be found using the Pythagorean theorem in the context of a two-dimensional vector. For a vector \( v = \langle a, b \rangle \), the magnitude is given by the formula:

• \[ \|v\| = \sqrt{a^2 + b^2} \]

In our example with vector \( v = \langle 4, -5 \rangle \), you plug in the values for \( a \) and \( b \) as 4 and -5, respectively.
You then calculate:

• \( \|v\| = \sqrt{4^2 + (-5)^2} = \sqrt{16 + 25} = \sqrt{41} \)

This result \( \sqrt{41} \) is the "length" of the vector, showing how large the vector is in terms of its magnitude.

Vector Direction

Vector direction tells you where the vector is pointing. It's all about the orientation of the vector in space. Imagine an arrow pointing somewhere; the angle or path it shows is the direction. For our vector \( v = \langle 4, -5 \rangle \), the direction is defined by the components 4 and -5, meaning it points right and down in a 2D coordinate system.
The direction becomes more interesting when calculating a unit vector, aiming to keep the same direction but with a magnitude of 1.
Unit vectors help standardize direction, showing us a way to direct the vector without worrying about its length. We find them by dividing each component of the vector by the vector's magnitude.

Vector Components

Vector components are simply the building blocks of a vector, representing how much "influence" a vector has in each direction. In a 2D space, like our vector \( v = \langle 4, -5 \rangle \), the components are the \( x \) direction (horizontal) component 4 and the \( y \) direction (vertical) component -5.

• The first number, 4, tells us the vector moves 4 units in the positive \( x \)-direction.
• The second number, -5, means it moves 5 units in the negative \( y \)-direction.

These components are critical when converting a vector to a unit vector. By understanding each part of the vector, you can normalize it, making it a unit vector while preserving its direction.
Once you know the magnitude, dividing the original components by this magnitude:

• \( u = \langle \frac{4}{\sqrt{41}}, \frac{-5}{\sqrt{41}} \rangle \)

creates a vector with the same direction, now standardized to unit length. This process allows us to keep the important directional information intact while adjusting the vector's magnitude.