Problem 66 - Trigonometry

Question

Approximate the magnitude of each vector and the angle \(\theta, 0^{\circ} \leq \theta<360^{\circ}\), that the vector makes with the positive \(x\)-axis. Round your answers to the nearest tenth. $$F=-6 i-\sqrt{3} \mathbf{j}$$

Step-by-Step Solution

Answer

The magnitude is approximately 6.2, and the angle is approximately 196.1°.

Step 1: Identify the Components

The given vector is \( F = -6i - \sqrt{3}j \). Identify the components: \( a = -6 \) (component along the x-axis) and \( b = -\sqrt{3} \) (component along the y-axis).

Step 2: Calculate the Magnitude

The magnitude \( |F| \) of the vector is calculated using the formula \( |F| = \sqrt{a^2 + b^2} \). Substitute the values: \( |F| = \sqrt{(-6)^2 + (-\sqrt{3})^2} = \sqrt{36 + 3} = \sqrt{39} \). Approximating, we get \( |F| \approx 6.2 \).

Step 3: Determine the Angle with Positive x-axis

The angle \( \theta \) that the vector makes with the positive \( x \)-axis is given by \( \theta = \arctan\left(\frac{b}{a}\right) \). Substitute \( a = -6 \) and \( b = -\sqrt{3} \): \( \theta = \arctan\left(\frac{-\sqrt{3}}{-6}\right) \approx \arctan\left(\frac{\sqrt{3}}{6}\right) \approx 16.1^\circ \).

Step 4: Adjust the Angle to Correct Quadrant

Since both components are negative, the vector is in the third quadrant. Therefore, to find the correct direction, add \( 180^\circ \) to the angle from step 3: \( \theta \approx 16.1^\circ + 180^\circ = 196.1^\circ \).

Key Concepts

Vector Magnitude

Understanding how to find the magnitude of a vector is crucial in many fields of science and engineering. The magnitude of a vector, often represented as \(|F|\), gives you a sense of how long the vector is. Imagine this as the distance from the starting point of the vector to its endpoint.
To calculate the magnitude, use the formula for a vector with components \(a\) and \(b\):

• \(|F| = \sqrt{a^2 + b^2}\)

By plugging in our values, \(a = -6\) and \(b = -\sqrt{3}\), the magnitude becomes \(|F| = \sqrt{36 + 3} = \sqrt{39}\), which is approximately \(6.2\) when rounded to the nearest tenth.
This tells us that the vector stretches about \(6.2\) units in space, regardless of its direction.

Vector Angle

Angles in vectors provide information about the direction in which the vector is pointing. Specifically, the angle \(\theta\) informs us about the tilt of the vector with respect to the positive x-axis. To find this angle, we use the arctangent function:

• \(\theta = \arctan\left(\frac{b}{a}\right)\)

Substituting our components \(a = -6\) and \(b = -\sqrt{3}\) gives us \(\theta = \arctan\left(\frac{-\sqrt{3}}{-6}\right)\), which simplifies to \(\arctan\left(\frac{\sqrt{3}}{6}\right)\). Calculating this gives an approximate angle of \(16.1^\circ\).
However, because both \(a\) and \(b\) are negative, our vector is actually pointing to the third quadrant. This means we need to adjust our angle by adding \(180^\circ\), resulting in a final angle \(\theta\) of approximately \(196.1^\circ\). This adjustment ensures that the angle accurately reflects the vector's position in relation to the x-axis.

Trigonometric Functions

Trigonometry is at the heart of understanding vector directions and magnitudes. One of the main tools used is the arctangent function, denoted as \(\arctan\). This function helps determine the angle \(\theta\) based on the ratio of the vector's y-component \(b\) to its x-component \(a\).

• Trigonometric functions like sine, cosine, and tangent help relate the sides and angles of triangles formed by vectors.
• The arctangent function is specifically useful for finding angles when you have two sides of a right triangle.

In our vector problem, the arctangent allows us to "reverse-engineer" the angle using \(\frac{b}{a}\). Once calculated, this initial angle must often be adjusted based on which quadrant the vector lies.
Remember, trigonometric functions provide the mathematical structure that explains why vectors point where they do, based on their components.