91Ó°ÊÓ

Trigonometric functions like \(\text{sin}\) and \(\text{cos}\) are essential when working with vector components. These functions help convert between the angle of a vector and its direction in the coordinate system. In our example, we have the direction angle of \(\theta = 250^{\text{\textdegree}}\). The key functions are: \(\text{cos} \theta\) gives the ratio of the adjacent side (x-component) to the hypotenuse (magnitude of the vector), while \(\text{sin} \theta\) gives the ratio of the opposite side (y-component) to the hypotenuse.
To apply these:

• We found \(a_x\) using \(a_x = a \text{cos} \theta\), where \(a\) is the magnitude and \( \theta \) is the angle.
• Similarly, we found \(a_y\) using \(a_y = a \text{sin} \theta\).

These trigonometric functions simplify calculations and make it easier to find vector components when the angle and magnitude are known.

The magnitude of a vector represents its length and is a critical part of determining its components. It's the distance from the origin to the point in the coordinate system where the vector ends.
For our problem, we have a vector \(\text{\textbar}\textbf{a}\text{\textbar} = 7.3 \text{ m}\). The components' calculation relies on this magnitude. The formulas \(a_x = a \text{cos} \theta\) and \(a_y = a \text{sin} \theta\) both include the magnitude \(a\). This means any changes in the magnitude directly affect both the x and y components.
Consider that the magnitude helps convert trigonometric functions, typically ratios, into actual distances or lengths in the coordinate system. Always ensure the magnitude is properly understood before proceeding with component calculations.

Understanding the coordinate system is fundamental when working with vectors. It provides a framework to describe the vector’s direction and magnitude. A standard coordinate system places vectors in the xy-plane with two perpendicular axes - the x-axis (horizontal) and the y-axis (vertical).
In this problem, we deal with the vector \( \textbf{a} \) in the xy-plane with an angle \(\theta \) measured counterclockwise from the positive x-axis. A key step was converting the angle to its standard position. Since \(250^{\text{\textdegree}}\) is in the third quadrant:

• Subtract \(180^{\text{\textdegree}}\) to find the reference angle: \(250^{\text{\textdegree}} - 180^{\text{\textdegree}} = 70^{\text{\textdegree}}\).

This helps identify the correct trigonometric values for the x and y components. Understanding which quadrant your vector is in ensures you get the right signs (positive or negative) for components. In our example, both components are negative since the vector is in the third quadrant where both x and y are negative.