Problem 39 - 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 . $$\mathbf{u}=\langle 3,0\rangle$$

Step-by-Step Solution

Answer

The unit vector in the direction of the given vector \( \mathbf{u}=\langle 3,0\rangle \) is \( \hat{\mathbf{u}} = \langle 1,0\rangle \) and its magnitude is indeed 1.

Step 1: Finding the Magnitude of the Vector

The magnitude of vector \( \mathbf{u}=\langle 3,0\rangle \) is obtained by the formula \( ||\mathbf{u}|| = \sqrt{u_x^2 + u_y^2} \). Substituting gives \( ||\mathbf{u}|| = \sqrt{3^2 + 0^2} = 3 \).

Step 2: Finding the Unit Vector

The unit vector \( \hat{\mathbf{u}} \) in the direction of \( \mathbf{u} \) is obtained by dividing vector \( \mathbf{u} \) by its magnitude. Therefore, \( \hat{\mathbf{u}} = \frac{\mathbf{u}}{||\mathbf{u}||} = \frac{\langle 3,0\rangle}{3} = \langle 1,0\rangle \).

Step 3: Verify the Magnitude of the Unit Vector

The magnitude of the unit vector \( \hat{\mathbf{u}} = \langle 1,0\rangle \) must equal 1. Using the formula \( ||\hat{\mathbf{u}}|| = \sqrt{u_x^2 + u_y^2} \), substituting gives \( ||\hat{\mathbf{u}}|| = \sqrt{1^2 + 0^2} = 1 \). Thus, the result is verified.

Key Concepts

Vector Magnitude

The concept of vector magnitude is fundamental to understanding vectors. The magnitude of a vector is essentially its length or size. To calculate this, we use the formula: \[ ||\mathbf{u}|| = \sqrt{u_x^2 + u_y^2} \] where

• \( u_x \) and \( u_y \) are the components of the vector along the x and y axes, respectively.

Here, for the vector \( \mathbf{u}=\langle 3,0\rangle \), we substitute the values into the formula:
\[ ||\mathbf{u}|| = \sqrt{3^2 + 0^2} = 3 \] This tells us that the vector has a magnitude or length of 3 units. This value is crucial as it helps in finding the direction and unit vector associated with \( \mathbf{u} \). Understanding vector magnitude ensures that we can accurately describe the size of vectors in various applications.

Direction of a Vector

Understanding the direction of a vector is just as important as knowing its magnitude. The direction is determined by the position of the vector in space, typically given as the angle it makes with the reference axis or as a unit vector.
A unit vector gives a clear direction without size, as it maintains the direction of the original vector but has a magnitude of 1. To find it, you divide the vector by its magnitude. The formula for finding a unit vector \( \hat{\mathbf{u}} \) in the direction of vector \( \mathbf{u} \) is:
\[ \hat{\mathbf{u}} = \frac{\mathbf{u}}{||\mathbf{u}||} \] For our vector \( \mathbf{u}=\langle 3,0\rangle \), the unit vector calculation is straightforward:

• Divide each component of the vector by its magnitude:
• \[ \hat{\mathbf{u}} = \frac{\langle 3,0\rangle}{3} = \langle 1,0\rangle \]

This result \( \langle 1,0\rangle \) tells us the direction of the vector lies entirely along the positive x-axis, confirming its directional quality without affecting size.

Magnitude Verification

Magnitude verification is a crucial step to ensure that a unit vector is correctly identified. A unit vector must have a magnitude of 1. To verify this, use the magnitude formula on \( \hat{\mathbf{u}} \):
\[ ||\hat{\mathbf{u}}|| = \sqrt{u_x^2 + u_y^2} \] For the unit vector \( \hat{\mathbf{u}} = \langle 1,0\rangle \), substitute its components into the formula:

• \( ||\hat{\mathbf{u}}|| = \sqrt{1^2 + 0^2} = \sqrt{1} = 1 \)

This confirms that the magnitude of our unit vector is indeed 1, hence verifying the result. It is vital to perform magnitude verification to ensure results are accurate, particularly when aligning with the definition of unit vectors.