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When working with vectors, one of the initial steps is to understand their components. A vector in a two-dimensional plane can be broken down into two main parts: the x-component and the y-component. These components can be thought of as the shadows or projections of the vector along the x-axis and y-axis.
For the displacement vector \(\overrightarrow{\mathbf{r}}\), the given components are:\

• The x-component has a magnitude of 125 meters and points along the negative x-axis. This means that if you were to draw it, the arrow would extend 125 meters to the left (since negative x-direction is usually to the left on a standard graph).
  
• The y-component has a magnitude of 184 meters and points along the negative y-axis, directing 184 meters downward on the graph.
  

Understanding these pieces helps in visualizing how the vector behaves and sets the stage for further calculations regarding its overall magnitude and direction.

Finding the magnitude of a vector is like finding the length of the hypotenuse in a right triangle. This is because the x and y components act as legs of a right triangle, with the vector itself as the hypotenuse. To compute the magnitude of the vector \(\overrightarrow{\mathbf{r}}\), we use the Pythagorean theorem:
\[|\overrightarrow{\mathbf{r}}| = \sqrt{x^2 + y^2} = \sqrt{(125)^2 + (184)^2}\]Breaking down the mathematics:

• \((125)^2\) equals 15625
• \((184)^2\) equals 33856

Adding these together gives us 49481, and taking the square root yields approximately 222.41 meters for the magnitude of vector \(|\overrightarrow{\mathbf{r}}|\).
This magnitude represents the straight-line distance from the vector's starting point to its ending point, combining both the horizontal and vertical displacements.

Once the magnitude is known, determining the direction is the next step. The direction of a vector is often expressed as an angle, often calculated using inverse trigonometric functions. For vector \(\overrightarrow{\mathbf{r}}\), we determine the angle \(\theta'\) with respect to the negative x-axis using the tangent function:
\[\tan(\theta') = \frac{|y|}{|x|} = \frac{184}{125}\]Calculating the inverse tangent (\(\tan^{-1}\)) of this ratio gives approximately 55.3 degrees.
However, because both components point in negative directions (third quadrant on a coordinate plane), the angle \(\theta\) with respect to the negative x-axis is adjusted:
\[\theta = 180^\circ - 55.3^\circ = 124.7^\circ\]This angle indicates that the vector \(\overrightarrow{\mathbf{r}}\) forms an angle of 124.7 degrees with the direction of the negative x-axis, showing the path it follows in its particular quadrant positioning.