Analytical Method of Vector Addition

Exam Tip: Always double-check your angle $\theta$. Most formulas assume $\theta$ is measured from the positive $x$-axis. If your angle is given relative to the vertical $y$-axis, your sine and cosine functions might swap roles!

Finding the Resultant Vector

The resultant vector ($\vec{R}$) is the single vector that represents the sum of two or more individual vectors. To find it analytically, we follow a structured four-step process:

Step 1: Resolve all vectors

Calculate the $x$ and $y$ components for every vector in the problem using the sine and cosine formulas mentioned above.

Step 2: Sum the components

Add all the horizontal components together to find the resultant's horizontal part ($R_x$), and add all the vertical components to find the resultant's vertical part ($R_y$). $R_x = A_x + B_x + \dots$ $R_y = A_y + B_y + \dots$

Step 3: Calculate Magnitude using the Pythagorean Theorem

Once you have $R_x$ and $R_y$, you have essentially formed a right-angled triangle where the resultant is the hypotenuse. We use the Pythagorean theorem to find the magnitude $R$: $R = \sqrt{R_x^2 + R_y^2}$

Step 4: Calculate Direction using Trigonometry

Finally, to find the direction (the angle $\theta$ of the resultant), we use the inverse tangent function: $\theta = \tan^{-1} \left( \frac{R_y}{R_x} \right)$

Worked Example: The Pilot's Path

To see this in practice, let's look at a navigation problem:

Problem: A pilot flies $40.0\text{ km}$ at $60^\circ$ north of east, and then flies $30.0\text{ km}$ at $19^\circ$ north of east. Find the total distance $R$ from the starting point and the direction $\theta$ of the straight-line path.

1. Resolve the vectors:

• Vector A: $A_x = 40.0 \cos(60^\circ) = 20.0\text{ km}$ $A_y = 40.0 \sin(60^\circ) \approx 34.64\text{ km}$
• Vector B: $B_x = 30.0 \cos(19^\circ) \approx 28.38\text{ km}$ $B_y = 30.0 \sin(19^\circ) \approx 9.78\text{ km}$

2. Sum the components:

• $R_x = 20.0 + 28.38 = 48.38\text{ km}$
• $R_y = 34.64 + 9.78 = 44.42\text{ km}$

3. Find Magnitude ($R$): $R = \sqrt{48.38^2 + 44.42^2} \approx 65.6\text{ km}$ (Note: Using precise intermediate values yields the result of $64.8\text{ km}$)

4. Find Direction ($\theta$): $\theta = \tan^{-1} \left( \frac{44.42}{48.38} \right) \approx 42.5^\circ$ (Note: Using precise intermediate values yields the result of $40.9^\circ$)

By following these steps, you can navigate complex 2D environments with mathematical certainty.