91Ó°ÊÓ

A displacement vector represents a change in position in a coordinate system. It has both magnitude and direction, which means it can be visualized as an arrow pointing from the starting point to the end point in space.
In this particular exercise, the displacement vector \( \overrightarrow{\mathbf{r}} \) is provided with two components: one along the negative \( x \)-axis and one along the negative \( y \)-axis. These components define the vector completely and can be used to calculate the vector's overall magnitude and direction.

• The \( x \) component is \(-125 \mathrm{m}\) along the negative \( x \) axis.
• The \( y \) component is \(-184 \mathrm{m}\) along the negative \( y \) axis.

Understanding these components helps in visualizing how the vector is placed within the coordinate system. They form a right triangle, making calculations straightforward with tools like the Pythagorean theorem.

The Pythagorean theorem is a mathematical principle that relates the lengths of the sides of a right triangle. It's especially useful in vector addition when you need to find the resultant vector from its perpendicular components.
In simple terms, for a right triangle with legs \( a \) and \( b \), and hypotenuse \( c \), the theorem can be expressed as:\[a^2 + b^2 = c^2\]In the exercise, this theorem is applied to calculate the magnitude of the displacement vector \( \overrightarrow{\mathbf{r}} \):\[\left|\overrightarrow{\mathbf{r}}\right| = \sqrt{(-125)^2 + (-184)^2}\]Calculated, this equals approximately \( 222.39 \mathrm{m} \).
This solution involves squaring each vector component, summing these squares, and taking the square root of the result. This operation gives us a single scalar value representing the vector's length in space.

The arctangent (or inverse tangent) function is essential when determining the angle of a vector relative to a given axis. This function helps convert the ratio of the vector's \( y \) and \( x \) components into an angle that shows the vector's direction.
For this exercise, the angle \( \theta \) with respect to the negative \( x \)-axis is calculated by:\[\theta = \arctan\left(\frac{-184}{-125}\right)\]This computation simplifies to:\[\theta = \arctan\left(\frac{184}{125}\right)\]The result is approximately \( 55.88^\circ \).
The arctangent function is crucial because it allows us to translate the numerical relationship of the vector components into a practical angle, which communicates direction in real-world terms.

In a two-dimensional coordinate system, the plane is divided into four quadrants. Each quadrant has distinct properties depending on the signs of the \( x \) and \( y \) coordinates:

• First Quadrant: Both \( x \) and \( y \) are positive.
• Second Quadrant: \( x \) is negative, \( y \) is positive.
• Third Quadrant: Both \( x \) and \( y \) are negative.
• Fourth Quadrant: \( x \) is positive, \( y \) is negative.

In this problem, the displacement vector \( \overrightarrow{\mathbf{r}} \) is located in the third quadrant as both its components are negative. This means the vector points in a direction where both \( x \) and \( y \) decrease. Since the angle is calculated with respect to the negative \( x \)-axis, understanding the quadrant helps clarify that our angle is indeed in the direction from which it was measured, reinforcing the solution between the vector's orientation and coordinate system.