Problem 62 - Trigonometry

Question

Find the magnitude of each vector and the angle \(\theta, 0^{\circ} \leq \theta<360^{\circ}\), that the vector makes with the positive \(x\)-axis. $$\mathbf{F}=-2 \sqrt{3} \mathbf{i}+2 \mathbf{j}$$

Step-by-Step Solution

Answer

Magnitude is 4, and the angle is \(150^{\circ}\).

Step 1: Identify the Components

The vector \( \mathbf{F} = -2 \sqrt{3} \mathbf{i} + 2 \mathbf{j} \) has components: \( x = -2 \sqrt{3} \) and \( y = 2 \).

Step 2: Calculate the Magnitude

Use the magnitude formula for vectors: \( ||\mathbf{F}|| = \sqrt{x^2 + y^2} \). Substituting with the given values, we have:\[||\mathbf{F}|| = \sqrt{(-2 \sqrt{3})^2 + (2)^2} = \sqrt{12 + 4} = \sqrt{16} = 4\]So, the magnitude of the vector is 4.

Step 3: Calculate the Angle with the Positive x-axis

Use \( \tan \theta \) to find the angle \( \theta \) with: \( \tan \theta = \frac{y}{x} \). Substituting the values:\[\tan \theta = \frac{2}{-2 \sqrt{3}} = -\frac{1}{\sqrt{3}}\]The angle whose tangent is \(-\frac{1}{\sqrt{3}}\) is \(-30^{\circ}\). However, since \( x < 0 \) and \( y > 0 \), the vector lies in the second quadrant. So, we adjust \( -30^{\circ} \) to its equivalent angle in the second quadrant:\[\theta = 180^{\circ} - 30^{\circ} = 150^{\circ}\]Thus, the angle \( \theta \) is \(150^{\circ}\).

Key Concepts

Vector Components

When working with vectors, it's essential to understand that they are essentially described by two main components in a 2D Cartesian plane. These components are usually along the x-axis and y-axis, hence they're often referred to as the x-component and y-component. In our specific case, the vector \( \mathbf{F} = -2 \sqrt{3} \mathbf{i} + 2 \mathbf{j} \) has:

• x-component: \(-2 \sqrt{3}\)
• y-component: \(2\)

Each component is crucial because it represents the vector's influence in that particular direction. To better imagine this, think of each component stretching the vector along its particular axis. This breakdown allows us to manipulate and understand the vector's behavior quantitatively.

Magnitude Formula

The magnitude of a vector is somewhat similar to the length of a line segment. It's a measure of how long the vector is, in a geometric sense. To find it, we use the Pythagorean theorem-like formula given by:\[||\mathbf{F}|| = \sqrt{x^2 + y^2}\]For our vector, substituting the x and y components:

• \(x = -2 \sqrt{3}\)
• \(y = 2\)

The formula becomes:\[||\mathbf{F}|| = \sqrt{(-2 \sqrt{3})^2 + (2)^2} = \sqrt{12 + 4} = \sqrt{16} = 4\]This indicates our vector's "length" is 4, regardless of direction. The magnitude is always non-negative and helps compare vector sizes.

Angle Calculation

To determine the angle that a vector makes with the positive x-axis, we often use tangent relationships, specifically the arctan or inverse tangent function. The basic idea is expressed in the formula:\[\tan \theta = \frac{y}{x}\]Plugging in the components of our vector \(y = 2\) and \(x = -2 \sqrt{3}\):\[\tan \theta = \frac{2}{-2 \sqrt{3}} = -\frac{1}{\sqrt{3}}\]This value corresponds to an angle of \(-30^{\circ}\). However, angles can be tricky because vectors can reside in different quadrants, and each affects the final angle differently. In this particular scenario, since the vector's x-component is negative and the y-component is positive, our vector is in the second quadrant.Thus, we adjust the result to reflect its actual position:\[\theta = 180^{\circ} - 30^{\circ} = 150^{\circ}\]Through this understanding, you can see how quadrants affect angle calculations!

Trigonometric Functions

Trigonometric functions play a pivotal role in vector analysis, mainly because they allow us to break down or reconstruct angles and components. Here, we primarily dealt with the tangent function, \(\tan \), which links our vector's components to its angle.Let's quickly break down these essential functions:

• Sine and Cosine relate to y and x-axis projections of the vector:
  • \(\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}\)
  • \(\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}\)
• Tangent is the ratio of sine and cosine, which gives angles as \(\frac{y}{x}\).

Knowing how these relationships work empowers you to solve vector problems by simplifying complex geometric relationships into understandable terms. This foundation is core to not just vectors, but wider physics and engineering applications.