Question

Vector \(\overrightarrow{\mathbf{a}}\) has a magnitude of \(5.2\) units and is directed east. Vector \(\overrightarrow{\mathbf{b}}\) has a magnitude of \(4.3\) units and is directed \(35^{\circ}\) west of north. By constructing vector diagrams, find the magnitudes and directions of (a) \(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}\), and \((\mathrm{b}) \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}\).

Short answer

The magnitudes and directions of \( \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}} \) and \( \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}} \) can be obtained by converting the vectors from polar to Cartesian form, performing vector addition and subtraction, and converting the results back to polar form.

Step-by-step solution

Convert Polar to Cartesian Form

For a vector directed east and west, this is similar to positive and negative x-direction respectively. And a vector directed north equals positive y-direction. So, convert polar vectors \( \overrightarrow{\mathbf{a}} \) and \( \overrightarrow{\mathbf{b}} \) to Cartesian form. \( \overrightarrow{\mathbf{a}} = 5.2\hat{i} \) (as it is directed towards the east) and \( \overrightarrow{\mathbf{b}} = 4.3cos(35^{\circ})(-\hat{i}) + 4.3sin(35^{\circ})\hat{j} \) (as it is directed 35 degrees west of north).

Perform Vector Addition and Subtraction

Calculate \( \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}} \) and \( \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}} \) using the Cartesian vectors obtained from the previous step.

Convert Resultant Vectors Back to Polar Form

Convert \( \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}} \) and \( \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}} \) back to their polar forms to find the magnitude and direction of the resultant vectors. The magnitude is given by the formula \( \sqrt{x^{2}+y^{2}} \) and the direction by \( \text{arctan}(y/x) \), where x and y are the components of the resultant vectors in the Cartesian plane.

Key concepts

Polar to Cartesian Conversion

Vectors can be expressed in different forms, and understanding these forms is crucial when dealing with vector mathematics. **Polar coordinates** represent vectors using a magnitude and an angle. For example, a vector might be described as having a length of 5 units and an angle of 30 degrees from a reference direction. But sometimes, it's more helpful to convert this representation into a Cartesian coordinate system which uses x and y components to describe the vector.

The conversion from polar to Cartesian is done using trigonometry:

• The x-component is found by multiplying the magnitude by the cosine of the angle: \(x = r \cos(\theta)\).
• The y-component is found by multiplying the magnitude by the sine of the angle: \(y = r \sin(\theta)\).

For \( \overrightarrow{\mathbf{b}} \), which is directed 35 degrees west of north, west is negative x and north is positive y. We calculate \(x\) and \(y\) as follows:
- \(x = 4.3 \cos(35^{\circ})(-1)\) - \(y = 4.3 \sin(35^{\circ})\)

This conversion allows us to perform arithmetic operations on vectors easily.

Vector Components

Understanding the components of a vector is like understanding its building blocks in a Cartesian system. Each vector can be broken down into two main parts: the **x-component** and the **y-component**. These components tell us how far the vector stretches along the x-axis and y-axis respectively.

Consider vector \( \overrightarrow{\mathbf{a}} = 5.2\hat{i}\). This means it stretches 5.2 units along the x-axis, with no movement in the y-direction (i.e., 0\hat{j}).
On the other hand, for \( \overrightarrow{\mathbf{b}}\), we found the components from its polar version:
- \(x = 4.3 \cos(35^{\circ})(-1),\) as it goes against the east direction- \(y = 4.3 \sin(35^{\circ})\).

To add or subtract vectors, simply add or subtract these components separately. If given vectors \(\underline{\mathbf{v_1}} = (x_1, y_1)\) and \(\underline{\mathbf{v_2}} = (x_2, y_2)\), the sum is \(\underline{\mathbf{v_1}} + \underline{\mathbf{v_2}} = (x_1 + x_2, y_1 + y_2)\).
Breaking vectors down in this way simplifies calculations and reveals the underlying geometric scenarios.

Magnitude and Direction Calculation

Once you have the components of a vector, you can easily find its **magnitude** and **direction**. This is a re-conversion to polar form to understand how long the resultant vector is and which direction it points.

To find the magnitude of a vector with components \(x\) and \(y\), use the Pythagorean theorem: \[|\underline{\mathbf{v}}| = \sqrt{x^2 + y^2}\]
This gives the vector's length, similar to finding the hypotenuse of a right triangle.

For the direction, utilize the tangent function to find the angle \(\theta\) the vector makes with the x-axis:
\[\theta = \text{arctan} \left( \frac{y}{x} \right)\]

This angle defines the vector's direction in a standard polar coordinate framework. Remember, depending on which quadrant your resultant vector lies in, you might need to adjust the angle for correct directional interpretation.
Combining these two measurements allows you to fully understand the vector's orientation and size, completing your vector operation analysis.