91Ó°ÊÓ

Vectors are quantities that have both magnitude and direction. In problems like this, it's important to represent vectors in a way that breaks them down into their component parts. This makes calculations more straightforward.

For example, in the given exercise, vector \( \overrightarrow{\mathbf{A}} \) is aligned with the positive \( y \)-axis, and can be written as \( 0 \hat{i} + 12 \hat{j} \) meters. This means it has no movement in the \( x \)-direction, hence the 0 \( \hat{i} \), and a movement of 12 meters in the \( y \)-direction, represented by 12 \( \hat{j} \).

Similarly, vector \( \overrightarrow{\mathbf{B}} \) points in the negative \( x \) direction, written as \( -33 \hat{i} + 0 \hat{j} \) meters. Here, -33 \( \hat{i} \) indicates the vector moves 33 meters in the opposite direction of the positive \( x \)-axis, and has no movement in the \( y \)-direction.

By breaking vectors down into components, we simplify addition and subtraction, helping us find resultant vectors more easily.

The magnitude of a vector is essentially its length or size. It tells us how much of something is represented by the vector. Calculating the magnitude involves using the Pythagorean theorem if the vector is split into its components.

For example, consider a vector \( \overrightarrow{R} = a \hat{i} + b \hat{j} \). The magnitude is found using \[ |\overrightarrow{R}| = \sqrt{a^2 + b^2} \].

In the exercise, the magnitude of each resultant vector, like \( \overrightarrow{R_a} = -33 \hat{i} + 12 \hat{j} \), is calculated with this formula: \( |\overrightarrow{R_a}| = \sqrt{(-33)^2 + (12)^2} \approx 35.1 \text{ m} \). The magnitude gives you the total size of the vector irrespective of its direction.

The direction of a vector shows us exactly where the vector is pointing. This is typically measured as an angle relative to a defined axis. In this exercise, we are often looking for the angle a vector makes with the \( x \)-axis.

To find this angle, trigonometric functions are useful. For a vector \( \overrightarrow{R} = a \hat{i} + b \hat{j} \), the direction \( \theta \) can be calculated by

• Using \( \tan(\theta) = \frac{b}{a} \)
• Then, \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \)

For instance, for \( \overrightarrow{R_a} = -33 \hat{i} + 12 \hat{j} \), the direction is \( \theta = \tan^{-1}\left(\frac{12}{-33}\right) \approx 20.2^\circ \) counterclockwise from the negative \( x \)-axis. Accurately determining the angle helps to fully describe the vector's direction.

The Pythagorean theorem is a key mathematical principle often used in calculating the magnitude of vectors. It states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.

Applied to vectors with components \( a \) and \( b \), you can determine the magnitude (hypotenuse) using the formula \[ c = \sqrt{a^2 + b^2} \].

In the context of the exercise, we see this principle applied multiple times, such as in calculating the magnitude of \( \overrightarrow{R_a} \) as \( \sqrt{1089 + 144} = \sqrt{1233} \approx 35.1 \text{ m} \). This theorem simplifies the process of finding how long a vector is, thus making it key to solving vector addition and subtraction problems.

Trigonometric functions such as sine, cosine, and tangent are vital when working with vectors. They help convert between angles and their vector components or vice versa.

In vector calculations, the tangent function is particularly useful for finding the angle a vector makes with an axis. For instance, \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \), letting us solve for \( \theta \) as \( \tan^{-1}\left(\frac{\text{opposite}}{\text{adjacent}}\right) \).

In the exercise, these concepts are illustrated clearly. For the resultant vector \( \overrightarrow{R_a} = -33 \hat{i} + 12 \hat{j} \), the angle \( \theta \) is derived from \( \tan(\theta) = \frac{12}{-33} \) giving \( \theta \approx 20.2^\circ \), providing its direction relative to an axis. Understanding and using these trigonometric functions make resolving vectors much easier.