91Ó°ÊÓ

Vectors are often expressed in unit-vector notation to clearly describe their direction and magnitude. Unit-vector notation uses standard basis vectors \( \hat{i} \) and \( \hat{j} \) which represent the x and y axes respectively. For example, a vector with components along the x and y axes can be written as \( \vec{A} = A_x \hat{i} + A_y \hat{j} \). Here, \( A_x \) and \( A_y \) are the magnitudes of the vector along the x and y axes. Using unit-vector notation helps simplify adding vectors since you can directly add their respective components.
In the exercise, we resolved each vector into its x and y components and summed them up as:
\( \vec{T} = 14.19 \hat{i} + 1.64 \hat{j} \).
This notation gives a clear understanding of the vector’s direction and magnitude in the 2D plane.

The magnitude of a vector tells us how long the vector is. Once we have the x and y components of the resultant vector, we can find its magnitude using the Pythagorean theorem. The formula for the magnitude \( T \) of a vector \( \vec{T} = x \hat{i} + y \hat{j} \) is:
\[ T = \sqrt{x^2 + y^2} \]
Using this formula, we sum the squares of the x and y components and then take the square root. By calculating:
\[ T = \sqrt{(14.19)^2 + (1.64)^2} \approx 14.29 \text{ m} \]
we find the magnitude of the resultant vector. The magnitude is an essential measure because it represents the total length of the path resulting from the addition of all individual vectors.

Finding the angle of a vector relative to a reference direction (often the positive x-axis) helps understand its orientation in space. To determine this angle, we use the inverse tangent function (arctangent), which relates the vector components to an angle:
\[ \theta = \tan^{-1}\bigg(\frac{y}{x}\bigg) \]
Given the x-component \( x \) and y-component \( y \) of the resultant vector, we can determine the angle it makes with the positive x-axis. For example:
\[ \theta = \tan^{-1}\bigg(\frac{1.64}{14.19}\bigg) \approx 6.57^\circ \]
This calculated angle tells us that the resultant vector is inclined at \( 6.57^\circ \) counterclockwise from the positive x-axis, providing crucial information about its direction.

Resolving vectors into their components is essential because it simplifies vector addition and other operations. A vector can always be broken down into its orthogonal (perpendicular) parts along the x-axis and y-axis. For instance, a vector \( \vec{P} \) with a magnitude of \( P \) at an angle \( \theta \) can be resolved as:
\[ P_x = P \cos(\theta) \]
\[ P_y = P \sin(\theta) \]
In the exercise, each vector was resolved into x and y components. By applying these formulas for each vector, we obtain:
For \( \vec{P} \),
\[ P_x = 10.0 \cos(25.0^\circ) \approx 9.06 \text{ m} \]
\[ P_y = 10.0 \sin(25.0^\circ) \approx 4.23 \text{ m} \]
Calculating the components simplifies the process of vector addition, making it straightforward to sum up all vectors.

Trigonometric functions, such as sine and cosine, are key to resolving vectors into components. These functions relate the angles with the ratios of the sides of a right triangle. When working with a vector at an angle \( \theta \):
\[ \cos(\theta) = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{x}{P} \]
\[ \sin(\theta) = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{y}{P} \]
These relations help find the x and y components of a vector. For example, for a vector of magnitude 10.0 m at an angle of 25.0°:
\[ P_x = 10.0 \cos(25.0^\circ) \approx 9.06 \text{ m} \]
\[ P_y = 10.0 \sin(25.0^\circ) \approx 4.23 \text{ m} \]
Mastering these functions is crucial for breaking down vectors and recomposing them accurately in other contexts, such as vector addition or finding resultant vectors.