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Displacement vectors are essential in physics as they represent the change in position of an object from its initial to its final point. These vectors have both magnitude and direction.
This makes them different from scalar quantities that only have magnitude. A displacement vector can be imagined as an arrow pointing from the starting position to the ending position.
The length of this arrow represents the magnitude, while the direction it points in indicates the direction of displacement.Understanding displacement vectors involves recognizing that they can be broken down into components along the Cartesian coordinate system, typically the x and y axes.
In the given exercise, we have three vectors,

• \(\overrightarrow{\mathbf{A}}\) with components \(A_x = 80.0 \, \mathrm{m}\) and \(A_y = 60.0 \, \mathrm{m}\)
• \(\overrightarrow{\mathbf{B}}\) with components derived from a magnitude \(B = 75.0 \, \mathrm{m}\) and \(B_x = 60.0 \, \mathrm{m}\)
• \(\overrightarrow{\mathbf{C}}\) defined by magnitude and angle

This decomposition into components helps in analyzing and comparing vectors more conveniently, as vectors that may otherwise be described at complex angles can be addressed in a simple orthogonal setup.

Vector components are crucial when dealing with vectors in two dimensions because they simplify complex problems into more manageable calculations along the x and y axes.
Any vector in a plane can be broken down into two perpendicular components. This is like taking a slanted line and finding how much it stretches horizontally and vertically.For vector \(\overrightarrow{\mathbf{A}}\) in this problem, we've already been given its components straightforwardly: \(A_x = 80.0 \, \mathrm{m}\) and \(A_y = 60.0 \, \mathrm{m}\).
The calculations for vector \(\overrightarrow{\mathbf{B}}\) are a bit more involved. Given its magnitude and one component, we used the Pythagorean theorem to find the other component:

• \(B_y = \sqrt{75.0^2 - 60.0^2} = 45.0 \, \mathrm{m}\)

For vector \(\overrightarrow{\mathbf{C}}\), which is given by magnitude and angle, trigonometry helps us find:

• \(C_x = 100.0 \cdot \cos(36.9^{\circ}) \approx 80.0 \, \mathrm{m}\)
• \(C_y = 100.0 \cdot \sin(36.9^{\circ}) \approx 60.0 \, \mathrm{m}\)

Breaking down vectors into these components makes it possible to compare and even add vectors by simply summing their respective x and y components.

Trigonometrical functions such as sine, cosine, and tangent are extremely useful tools in vector decomposition, particularly when vectors are defined by their angle and magnitude.
These functions allow us to find the horizontal (x) and vertical (y) components of a vector.For a vector at an angle \(\theta\) from the horizontal, the cosine function helps find the x-component, and the sine function helps find the y-component:

• \(x = r \cdot \cos(\theta)\)
• \(y = r \cdot \sin(\theta)\)

where \(r\) is the vector’s magnitude.
In vector \(\overrightarrow{\mathbf{C}}\), directed \(36.9^{\circ}\) above the positive x-axis and having a magnitude of \(100.0 \, \mathrm{m}\), we use trigonometry to determine:

• \(C_x = 100.0 \, \mathrm{m} \cdot \cos(36.9^{\circ}) \approx 80.0 \, \mathrm{m}\)
• \(C_y = 100.0 \, \mathrm{m} \cdot \sin(36.9^{\circ}) \approx 60.0 \, \mathrm{m}\)

Using trigonometric functions in this way simplifies the understanding of vectors and enhances the precision of calculations, which is indispensable in physics and engineering.