5.2 Vector Addition and Subtraction: Analytical Methods

Vector Addition and Subtraction

Vector addition and subtraction let you combine or compare quantities that have both magnitude and direction, like forces, velocities, and displacements. Instead of trying to add vectors graphically (drawing arrows tip-to-tail), the analytical method uses algebra and trigonometry to get exact answers. This is the approach you'll rely on for nearly every 2D problem going forward.

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Vector Components in Analytical Operations

Every vector in two dimensions can be broken into two pieces: a horizontal part and a vertical part. These are called components, and they're what make analytical vector math possible.

For a vector $\vec{A}$ with magnitude $A$ at angle $\theta$ from the positive x-axis:

• $A_x = A\cos\theta$ (the horizontal component, or projection onto the x-axis)
• $A_y = A\sin\theta$ (the vertical component, or projection onto the y-axis)

This process of splitting a vector into components is called vector decomposition. It works because any 2D vector can be fully described by how far it reaches in the x-direction and how far it reaches in the y-direction within a Cartesian coordinate system.

Why bother decomposing? Because once vectors are broken into components, you can use regular algebra on each axis separately. The x-components only interact with other x-components, and the same goes for y. This turns a tricky 2D problem into two straightforward 1D problems.

Vector Algebra and Unit Vectors

Unit vectors are vectors with a magnitude of exactly 1. They point along the coordinate axes and are written as $\hat{i}$ (x-direction), $\hat{j}$ (y-direction), and $\hat{k}$ (z-direction). Using unit vectors, you can write any vector in a compact form:

$\vec{A} = A_x\hat{i} + A_y\hat{j}$

This notation is just another way of saying "$A_x$ units in the x-direction and $A_y$ units in the y-direction." You'll see it used interchangeably with the ordered-pair notation $\vec{A} = (A_x, A_y)$.

Scalar multiplication means multiplying a vector by a plain number (a scalar). This changes the vector's magnitude but not its direction. For example, $2\vec{A}$ points the same way as $\vec{A}$ but is twice as long. Multiplying by a negative scalar reverses the direction.

Analytical Method for 2D Vector Math

Adding vectors: Add the corresponding components.

For vectors $\vec{A}$ and $\vec{B}$, the resultant $\vec{R} = \vec{A} + \vec{B}$ has components:

• $R_x = A_x + B_x$
• $R_y = A_y + B_y$

Example: If $\vec{A} = (3, 4)$ and $\vec{B} = (1, 2)$, then $\vec{R} = (3+1, 4+2) = (4, 6)$.

Subtracting vectors: Subtract the corresponding components.

For $\vec{R} = \vec{A} - \vec{B}$:

• $R_x = A_x - B_x$
• $R_y = A_y - B_y$

Example: If $\vec{A} = (5, 3)$ and $\vec{B} = (2, 1)$, then $\vec{R} = (5-2, 3-1) = (3, 2)$.

Vector subtraction is really just adding the negative of a vector. $\vec{A} - \vec{B}$ is the same as $\vec{A} + (-\vec{B})$, where $-\vec{B}$ has the same magnitude as $\vec{B}$ but points in the opposite direction.

Finding magnitude: Use the Pythagorean theorem on the components.

$|\vec{R}| = \sqrt{R_x^2 + R_y^2}$

Finding direction: Use the inverse tangent function.

$\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)$

where $\theta$ is measured from the positive x-axis. Watch out: your calculator's $\tan^{-1}$ only returns angles between $-90°$ and $90°$. If the resultant vector points into the second or third quadrant (negative $R_x$), you need to add $180°$ to get the correct angle. Always sketch the vector to confirm which quadrant it's in.

Vector Math in Physics Problem-Solving

Here's a reliable process for any problem involving multiple vectors:

1. Identify the vectors. Assign each one a symbol and determine its components. If you're given magnitude and angle, decompose using sine and cosine. For a projectile problem, your vectors might be initial velocity $\vec{v}_0$ and gravitational acceleration $\vec{g} = (0, -9.8)$ m/s².

2. Determine the operation. Think about what the problem is asking. Finding a net force? Add the force vectors. Finding the change in velocity? That's subtraction ($\Delta\vec{v} = \vec{v}_f - \vec{v}_i$).

3. Perform the operation component-by-component. Example: If displacement vectors are $\vec{d}_1 = (3, 4)$ m and $\vec{d}_2 = (2, -1)$ m, the resultant displacement is $\vec{d}_R = (3+2,\; 4+(-1)) = (5, 3)$ m.

4. Calculate magnitude and direction if the problem asks for them. For $\vec{d}_R = (5, 3)$:

   • Magnitude: $|\vec{d}_R| = \sqrt{5^2 + 3^2} = \sqrt{34} \approx 5.83$ m
   • Direction: $\theta = \tan^{-1}\left(\frac{3}{5}\right) \approx 31.0°$ above the positive x-axis

5. Interpret the result. Translate the math back into physics. That resultant $\vec{d}_R$ means the object ended up 5.83 m from its starting point, at an angle of about 31° above the horizontal. Always check that your answer makes physical sense: Is the magnitude reasonable? Does the direction point where you'd expect?