Problem 25 - College Trigonometry

Question

Find the component form of the vector \(\vec{v}\) using the information given about its magnitude and direction. Give exact values. \(\|\vec{v}\|=26 ;\) when drawn in standard position \(\vec{v}\) lies in Quadrant IV and makes an angle measuring \(\arctan \left(\frac{5}{12}\right)\) with the positive \(x\) -axis

Step-by-Step Solution

Answer

The vector's component form is \( \langle 24, -10 \rangle \).

Step 1: Identify the angle in Quadrant IV

In Quadrant IV, the angle with the positive x-axis is calculated by subtracting the given angle from 360°. Since the tangent gives \ \( \theta = \arctan \left( \frac{5}{12} \right) \), we know this angle is with respect to the positive x-axis when it's directly in quadrant I. Hence, when considering Quadrant IV, \( \theta\) itself will be sufficient as \(\arctan\) delivers an angle that already considers the direction in Quadrant I which would be the regular angle if it's not needing conversion to standard position.

Step 2: Calculate the x-component

In standard position, the x-component of a vector is determined by using \(\vec{v}_x = \| \vec{v} \| \cdot \cos(\theta) \), where \(\theta = \arctan \left( \frac{5}{12} \right)\). Substitute: \( \| \vec{v} \| = 26 \) and \( \cos(\theta) = \frac{12}{13} \) (since this comes from the sides of the triangle deriving from tangent):\\\( \vec{v}_x = 26 \cdot \frac{12}{13} = 24 \).

Step 3: Calculate the y-component

In standard position, the y-component of a vector is determined by using \(\vec{v}_y = \| \vec{v} \| \cdot \sin(\theta) \), where \(\theta = \arctan \left( \frac{5}{12} \right)\). Substitute: \( \| \vec{v} \| = 26 \) and \( \sin(\theta) = \frac{5}{13} \) (since this comes from the sides of the triangle deriving from tangent):\\\( \vec{v}_y = 26 \cdot \frac{5}{13} = 10 \). However, since we're in Quadrant IV, the y-component should be negative: \( \vec{v}_y = -10 \).

Step 4: Determine the component form of the vector

Combine the results from the previous steps to express the vector in component form. The vector form is \( \vec{v} = \langle \vec{v}_x, \vec{v}_y \rangle = \langle 24, -10 \rangle \).

Key Concepts

Magnitude and Direction

When describing vectors, two essential aspects are magnitude and direction. The magnitude of a vector represents its length or size. It's similar to how you might measure the length of a line segment. For example, given that our vector \( \vec{v} \) has a magnitude of 26, imagine this as a distance measurement on a plane.
Direction indicates where the vector points from its origin to its terminal position. Think of it like a compass showing the heading from north. The angle given, \( \theta = \arctan \left( \frac{5}{12} \right) \), tells us the direction. Even if we don't know it immediately as degrees, we understand its relative position to the axes.
In this exercise, the vector \( \vec{v} \) not only has an angle but also a clear position on the coordinate system in Quadrant IV. Quadrants help us determine which direction components are positive or negative. Here, the vector points towards the positive x-axis and the negative y-axis.

Trigonometric Functions

Trigonometric functions, like sine and cosine, allow us to break down vectors into components. These functions work based on right-angled triangles from the vector's angle.
Here's how they come in handy: the cosine of an angle in a right triangle equals the adjacent side over the hypotenuse. When \( \theta = \arctan \left( \frac{5}{12} \right) \), the adjacent side is 12 and the opposite is 5, given by the relation of tangent's opposite over adjacent sides.
Therefore, we calculate:

• \( \cos(\theta) = \frac{12}{13} \) because it's adjacent (12) over hypotenuse (13).
• \( \sin(\theta) = \frac{5}{13} \) for the opposite (5) over hypotenuse (13).

These derived ratios help us find the x and y components of the vector. With this understanding, putting these ratios in context of the vector’s magnitude gives us precise components.

Quadrant Analysis

Understanding quadrants is key for analyzing the orientation of vectors on a coordinate plane. The Cartesian plane is divided into four quadrants, each affecting the sign of x and y components differently.
In our problem, \( \vec{v} \) lies in Quadrant IV, where the x-component is positive and the y-component is negative. Quadrant IV signifies the vector direction goes towards right and down. This is because the x-values are positive and y-values are negative here.
Given the expressions:

• \( \vec{v}_x = 26 \cdot \cos(\theta) = 24 \)
• \( \vec{v}_y = 26 \cdot \sin(\theta) = -10 \)

The formulas show numerical results reflecting the quadrant's directional rules. It's the quadrant analysis that adjusts the sign of the y-component, giving us the correct directional sense—hence the negative y-value.