Problem 16 - Thomas Calculus

Question

Find the component form of the vector. The unit vector obtained by rotating the vector \(\langle 1,0\rangle 135^{\circ}\) counterclockwise about the origin

Step-by-Step Solution

Answer

The component form is \( \langle -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \rangle \).

Step 1: Understand the initial vector

The vector given is \( \langle 1, 0 \rangle \). This is a unit vector along the positive x-axis. Our task is to rotate this vector by \( 135^{\circ} \) counterclockwise to find the new vector.

Step 2: Apply the rotation formula

The rotation of a vector \( \langle x, y \rangle \) by an angle \( \theta \) is given by the formula:\[ x' = x \cos(\theta) - y \sin(\theta)\]\[ y' = x \sin(\theta) + y \cos(\theta)\]We'll apply this to our vector \( \langle 1, 0 \rangle \) with \( \theta = 135^{\circ} \).

Step 3: Calculate new x-component

Substitute the values into the x-component formula:\[x' = 1 \cdot \cos(135^{\circ}) - 0 \cdot \sin(135^{\circ}) = \cos(135^{\circ})\]Thus, \( x' = -\frac{\sqrt{2}}{2} \).

Step 4: Calculate new y-component

Substitute the values into the y-component formula:\[y' = 1 \cdot \sin(135^{\circ}) + 0 \cdot \cos(135^{\circ}) = \sin(135^{\circ})\]Thus, \( y' = \frac{\sqrt{2}}{2} \).

Step 5: Confirm the vector is a unit vector

To confirm the new vector is a unit vector, calculate its magnitude:\[ \sqrt{\left(-\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = \sqrt{1} = 1\]The magnitude is 1, confirming it is a unit vector.

Key Concepts

Understanding Trigonometric Functions

Trigonometric functions, including sine (\( \sin \theta \)) and cosine (\( \cos \theta \)), play a vital role in vector rotation. These functions are based on the relationships between the angles and sides of right triangles. When dealing with vector rotation, the angle \( \theta \) indicates how much the vector is rotated around the origin.

• The cosine function consistently relates to the adjacent side of an angle in a right triangle and its hypotenuse. \( \cos \theta \) helps us find the horizontal component of a rotated vector.
• The sine function is associated with the opposite side of an angle and the hypotenuse. \( \sin \theta \) aids in calculating the vertical component.

In vector rotation, these functions help determine the new position of vector components by reallocating the original x and y components via the rotation formulas. These formulas adjust the vector components based on how much it's rotated, providing a way to precisely determine the new vector orientations after rotation.

Exploring the Unit Vector

A unit vector is a vector with a magnitude of exactly one. Think of it as a pure direction with no specific length, just orientation. It's often used for simplistic calculations and transformations because it retains the direction without influencing the size. Unit vectors are generally represented in the form of:

• The x-direction: \( \hat{i} = \langle 1, 0 \rangle \)
• The y-direction: \( \hat{j} = \langle 0, 1 \rangle \)

In our exercise, the vector \( \langle 1,0 \rangle \) is along the positive x-axis and is indeed a unit vector. After any transformation or rotation, ensuring that the resulting vector remains a unit vector is crucial, as this confirms the transformation didn't alter the vector's 'unit-ness'. The calculation of magnitude using properties of trigonometric functions reassures that the rotated unit vector retains its magnitude of one.

Angle Measurement and Rotation

Angles are measured typically in degrees or radians, providing a sense of rotation direction and magnitude. A rotation involves shifting a point or a vector from its original position by an angular measure, like 135 degrees in this exercise, counterclockwise.

• Degrees are a common way to measure angles, where a full circle equals 360 degrees.
• Radians, another measure, relate the angle to the radius of a circle, comprising \(2\pi\) radians in a complete circle.

When rotating a vector by a specific angle, understanding the direction—clockwise or counterclockwise—is vital. In our case, the vector \(\langle 1, 0 \rangle \) is rotated 135 degrees counterclockwise, changing its position within the plane. By applying trigonometric functions, the new angle determines how the original components of the vector are affected and placed on a new position on the coordinate system. This operation preserves unit magnitude while creating a fresh directional exposure.