Problem 52 - College Trigonometry

Question

For the given vector \(\vec{v}\), find the magnitude \(\|\vec{v}\|\) and an angle \(\theta\) with \(0 \leq \theta<360^{\circ}\) so that \(\vec{v}=\|\vec{v}\|\langle\cos (\theta), \sin (\theta)\rangle\) (See Definition 11.8.) Round approximations to two decimal places. $$ \vec{v}=\langle-114.1,42.3\rangle $$

Step-by-Step Solution

Answer

The magnitude is approximately 121.67, and the angle is approximately 159.65°.

Step 1: Calculate the Magnitude

The magnitude of a vector \( \vec{v} = \langle v_x, v_y \rangle \) is given by the formula \( \| \vec{v} \| = \sqrt{v_x^2 + v_y^2} \). For the vector \( \langle -114.1, 42.3 \rangle \), we compute the magnitude as follows:\[\| \vec{v} \| = \sqrt{(-114.1)^2 + (42.3)^2} = \sqrt{13020.81 + 1788.29} = \sqrt{14809.10} \approx 121.67.\]

Step 2: Find the Angle with the Positive X-Axis

The angle \( \theta \) can be found using the tangent function, where \( \tan(\theta) = \frac{v_y}{v_x} \). Since the vector \( \vec{v} = \langle -114.1, 42.3 \rangle \), we first compute \( \theta_{ref} \) without considering the sign:\[\tan(\theta_{ref}) = \left|\frac{42.3}{-114.1}\right| = \frac{42.3}{114.1} \approx 0.3708.\]Find \( \theta_{ref} = \arctan(0.3708) \approx 20.35^{\circ} \).Since the vector is in the second quadrant (negative x, positive y), the actual angle \( \theta = 180^{\circ} - \theta_{ref} \approx 180^{\circ} - 20.35^{\circ} = 159.65^{\circ}. \)

Key Concepts

Vector Angle

A vector angle is an important concept when dealing with vectors, especially in a plane. When given a vector, it's crucial to determine the angle it makes with the positive x-axis. This angle helps us understand the direction that the vector is pointing towards.
The angle, often denoted as \( \theta \), can typically be calculated using trigonometric functions, especially with the tangent function. The formula used is \( \tan(\theta) = \frac{v_y}{v_x} \), where \( v_y \) and \( v_x \) are the y and x components of the vector respectively.
If the vector lies in a quadrant where either or both components are negative, you'll need to consider the signs to determine the true direction of the angle in relation to the standard cartesian plane.

• First quadrant: No adjustment, use \( \theta = \arctan(\tan(\theta)) \).
• Second quadrant: Use \( \theta = 180^\circ - \theta_{ref} \).
• Third quadrant: Use \( \theta = 180^\circ + \theta_{ref} \).
• Fourth quadrant: Use \( \theta = 360^\circ - \theta_{ref} \).

This ensures the angle \( \theta \) is correctly calculated within its range \( 0^\circ \leq \theta < 360^\circ \).

Vector Components

Vector components are like the building blocks of a vector. They help us break down a vector into two perpendicular directions, usually the x-direction and y-direction in a 2D plane. For a vector \( \vec{v} = \langle v_x, v_y \rangle \), the components \( v_x \) and \( v_y \) represent how far the vector stretches along the x and y axes respectively.
Essentially, any vector can be expressed with these two components. By understanding the components, calculations involving vectors like finding magnitude, direction or performing addition become more straightforward.
The x-component \( v_x \) is often found using \( \|\vec{v}\| \cos(\theta) \) and the y-component \( v_y \) through \( \|\vec{v}\| \sin(\theta) \). Here, \( \|\vec{v}\| \) is the magnitude and \( \theta \) is the angle.

• Helps in understanding direction and length of the vector.
• Important in physics for force and velocity calculations.

Breaking vectors down into components adds a lot of clarity and aids in visualizing the vector's direction and length.

Trigonometric Functions

Trigonometric functions are fundamental in vector analysis. They allow us to deal with angles and distances in a plane. The primary trigonometric functions often used are sine, cosine, and tangent:

• **Sine (\( \sin \))**: Relates the y-component to the vector’s magnitude as \( \sin(\theta) = \frac{v_y}{\|\vec{v}\|} \).
• **Cosine (\( \cos \))**: Relates the x-component to the vector’s magnitude as \( \cos(\theta) = \frac{v_x}{\|\vec{v}\|} \).
• **Tangent (\( \tan \))**: Relates y and x components as \( \tan(\theta) = \frac{v_y}{v_x} \).

These functions are invaluable for finding angles between vectors and their respective axes.
When calculating these functions, especially when using the inverses (\( \arcsin, \arccos, \arctan \)), we are able to find angles from the given components. This is crucial in tasks such as rotating vectors or projecting one vector onto another. Trigonometric functions simplify the understanding of geometry in vector spaces and make it easier to solve more complex vector-related problems.