Problem 15 - Thomas Calculus

Question

Find the component form of the vector. The unit vector obtained by rotating the vector \(\langle 0,1\rangle 120^{\circ}\) counterclockwise about the origin

Step-by-Step Solution

Answer

The component form of the unit vector is \( \langle -\frac{\sqrt{3}}{2}, -\frac{1}{2} \rangle \).

Step 1: Understand the Initial Vector

The original vector is given as \( \langle 0, 1 \rangle \), which means it points vertically upwards along the y-axis from the origin.

Step 2: Set Up Rotation Matrix

To rotate a vector counterclockwise by \( \theta = 120^{\circ} \), we use the rotation matrix: \[R = \begin{pmatrix} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{pmatrix} = \begin{pmatrix} \cos 120^{\circ} & -\sin 120^{\circ} \ \sin 120^{\circ} & \cos 120^{\circ} \end{pmatrix}\]

Step 3: Calculate Trigonometric Values

The trigonometric values are:- \( \cos 120^{\circ} = -\frac{1}{2} \)- \( \sin 120^{\circ} = \frac{\sqrt{3}}{2} \)

Step 4: Apply Rotation Matrix

Multiply the vector \( \langle 0, 1 \rangle \) by the rotation matrix:\[\begin{pmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2} \ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} (-\frac{1}{2})(0) + (-\frac{\sqrt{3}}{2})(1) \ (\frac{\sqrt{3}}{2})(0) + (-\frac{1}{2})(1) \end{pmatrix} = \begin{pmatrix} -\frac{\sqrt{3}}{2} \ -\frac{1}{2} \end{pmatrix}\]

Step 5: Verify the Vector is a Unit Vector

To check if the result is a unit vector, confirm the magnitude:\[\sqrt{\left(-\frac{\sqrt{3}}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1\]Since the magnitude is 1, the vector is a unit vector.

Key Concepts

Unit Vector

A unit vector is a vector that has a length or magnitude of exactly 1. This means it points in a specific direction without concern for its length, which simplifies calculations such as direction or rotation.
Unit vectors are often used in physics and engineering to represent directional components. For any vector, you can find a corresponding unit vector by dividing each component of the vector by the vector's magnitude.
In our exercise, the resulting vector from the rotation \( \langle -\frac{\sqrt{3}}{2}, -\frac{1}{2} \rangle \) was proven to maintain a magnitude of 1, confirming it as a unit vector.

Rotation Matrix

A rotation matrix is a tool used to rotate vectors in a coordinate plane. It consists of trigonometric function values arranged in a 2x2 matrix: \[R = \begin{pmatrix} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{pmatrix}\]The angle \( \theta \) represents the amount of rotation, usually in degrees or radians.
For our problem, \( \theta \) was 120 degrees, and the rotation matrix was constructed using those specific angle's trigonometric values. This matrix allows you to rotate any vector, which demonstrates how powerful and versatile rotation matrices can be in vector mathematics.
By multiplying the original vector \( \langle 0, 1 \rangle \) with the rotation matrix, a new vector orientation is achieved.

Trigonometric Values

Trigonometric values, specifically sine and cosine, play a vital role in calculating vector rotations. They are derived from right-angled triangles and are essential for determining the component values of rotated vectors.
For the 120-degree rotation in the exercise, the values are:

• \( \cos 120^{\circ} = -\frac{1}{2} \)
• \( \sin 120^{\circ} = \frac{\sqrt{3}}{2} \)

These values were placed into the rotation matrix to transform the vector. Understanding trigonometric values allows for accurate predictions of how a vector's components change with rotation.

Magnitude Calculation

Magnitude calculation determines the length of a vector, representing the distance from the origin without any directional influence. To ensure a vector remains a unit vector after transformations, we verify its magnitude.
Given a vector \( \langle a, b \rangle \), its magnitude \( |\mathbf{v}| \) is calculated as:\[|\mathbf{v}| = \sqrt{a^2 + b^2}\]In our rotated vector \( \langle -\frac{\sqrt{3}}{2}, -\frac{1}{2} \rangle \), checking\[ \sqrt{\left(-\frac{\sqrt{3}}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} = 1 \]showed the magnitude to be 1. This verification step is crucial to confirm a vector remains of unit length, proving correctness of rotational operations.