Vectors and Vector Operations

TL;DR

You'll learn to break vectors into components, add/subtract them, and use dot and cross products to find angles, projections, and torques. You'll practice these on real numeric problems the way they show up on exams. By the end, you can solve any "find the resultant" or "find the angle between" problem without hesitating.

1. The Mental Model

A vector is just a number with a direction attached — it tells you "how much" and "which way." Every vector operation (addition, dot product, cross product) is really a rule for combining "how much" and "which way" from two or more vectors into a new answer. The core idea: components turn geometry into ordinary arithmetic.

2. The Core Material

2.1 Components and Addition — the foundation for everything else

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Any 2D vector can be written as $\vec{A} = A_x\hat{i} + A_y\hat{j}$, where:

$$A_x = A\cos\theta, \qquad A_y = A\sin\theta$$

and $\theta$ is measured from the positive x-axis. Going backward, given components you find magnitude and direction with:

$$A = \sqrt{A_x^2 + A_y^2}, \qquad \theta = \tan^{-1}\left(\frac{A_y}{A_x}\right)$$

Why this matters: once vectors are in component form, addition is trivial — you just add matching components:

$$\vec{R} = \vec{A} + \vec{B} = (A_x+B_x)\hat{i} + (A_y+B_y)\hat{j}$$

No triangle-drawing, no law of cosines needed. This is the entire trick behind every "two forces act on an object, find the net force" problem.

2.2 The Dot Product — measures "how aligned" two vectors are

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$$\vec{A}\cdot\vec{B} = A_xB_x + A_yB_y + A_zB_z = AB\cos\theta$$

The dot product gives a scalar (just a number, no direction). It's zero when vectors are perpendicular, maximum positive when they point the same way, and negative when they point mostly opposite directions.

Two big uses on exams:
1. Finding the angle between vectors — rearrange to $\theta = \cos^{-1}\left(\frac{\vec{A}\cdot\vec{B}}{AB}\right)$.
2. Work done by a force: $W = \vec{F}\cdot\vec{d} = Fd\cos\theta$. This is why a force perpendicular to displacement does zero work.

2.3 The Cross Product — measures "how perpendicular" and gives a new vector

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$$\vec{A}\times\vec{B} = (A_yB_z - A_zB_y)\hat{i} - (A_xB_z - A_zB_x)\hat{j} + (A_xB_y - A_yB_x)\hat{k}$$

Magnitude: $|\vec{A}\times\vec{B}| = AB\sin\theta$. Direction: found with the right-hand rule (point fingers along $\vec{A}$, curl toward $\vec{B}$, thumb points along the result).

This shows up in torque ($\vec{\tau} = \vec{r}\times\vec{F}$) and magnetic force ($\vec{F} = q\vec{v}\times\vec{B}$). Unlike the dot product, the cross product is zero when vectors are parallel, and maximum when they're perpendicular — the opposite behavior of $\cos\theta$ vs $\sin\theta$.

flowchart TD
    A["Two vectors A and B given"] --> B{"What do you need?"}
    B -->|"Net vector result"| C["Break into components,
    add matching components"]
    B -->|"Scalar: angle, work, projection"| D["Use dot product
    A·B = AxBx + AyBy"]
    B -->|"Vector: torque, area, rotation axis"| E["Use cross product
    right-hand rule for direction"]
    C --> F["Recombine: magnitude = sqrt(Rx^2+Ry^2)"]
    D --> G["theta = cos^-1(A·B / AB)"]
    E --> H["magnitude = AB sin(theta)"]

3. Worked Example

Let's do the kind of multi-part problem that actually shows up on exams.

Problem: Vector $\vec{A}$ has magnitude 8.0 N at an angle of 30° above the x-axis. Vector $\vec{B}$ has magnitude 5.0 N at an angle of 120° above the x-axis.

(a) Find the resultant $\vec{R} = \vec{A} + \vec{B}$ in component form and as magnitude/angle.
(b) Find the angle between $\vec{A}$ and $\vec{B}$ using the dot product.
(c) Find $\vec{A}\times\vec{B}$ (treat both as lying in the xy-plane, so the result is along $\hat{k}$).

Part (a) — Components and addition

First, break each vector into x and y components:

$$A_x = 8.0\cos(30°) = 8.0(0.866) = 6.93\text{ N}$$
$$A_y = 8.0\sin(30°) = 8.0(0.500) = 4.00\text{ N}$$

$$B_x = 5.0\cos(120°) = 5.0(-0.500) = -2.50\text{ N}$$
$$B_y = 5.0\sin(120°) = 5.0(0.866) = 4.33\text{ N}$$

Add the components:

$$R_x = 6.93 + (-2.50) = 4.43\text{ N}$$
$$R_y = 4.00 + 4.33 = 8.33\text{ N}$$

Now convert back to magnitude and angle:

$$R = \sqrt{4.43^2 + 8.33^2} = \sqrt{19.6 + 69.4} = \sqrt{89.0} = 9.43\text{ N}$$

$$\theta = \tan^{-1}\left(\frac{8.33}{4.43}\right) = \tan^{-1}(1.88) = 62.0°$$

So $\vec{R} \approx 9.4$ N at 62° above the x-axis. Check it makes sense: the resultant angle (62°) sits between A's angle (30°) and B's angle (120°) — exactly what you'd expect when adding two vectors. If your answer fell outside that range, you'd know you made an arithmetic error.

Part (b) — Dot product for angle between vectors

$$\vec{A}\cdot\vec{B} = A_xB_x + A_yB_y = (6.93)(-2.50) + (4.00)(4.33)$$
$$= -17.3 + 17.3 = 0.0$$

Interesting — the dot product comes out to (essentially) zero! That means $\vec{A}$ and $\vec{B}$ are perpendicular. Let's sanity check directly: the angle between them is $120° - 30° = 90°$. Yes — matches perfectly. This is a good habit: when a dot product surprises you, check it against the raw angle difference.

Part (c) — Cross product

Since both vectors lie in the xy-plane ($A_z = B_z = 0$), the cross product simplifies to just a z-component:

$$(\vec{A}\times\vec{B})_z = A_xB_y - A_yB_x = (6.93)(4.33) - (4.00)(-2.50)$$
$$= 30.0 - (-10.0) = 40.0$$

So $\vec{A}\times\vec{B} = 40.0\,\hat{k}$ N² (magnitude 40 N², pointing out of the page).

Check using the alternate formula: $AB\sin\theta = (8.0)(5.0)\sin(90°) = 40.0(1) = 40.0$. Matches exactly, confirming both our component arithmetic and the fact these vectors are perpendicular (since $\sin 90° = 1$ gives the maximum possible cross product for these magnitudes).

4. Key Takeaways

4.1 Most Important Concepts

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• Components are the universal translator. Any vector problem becomes simple addition/subtraction once everything is in $(x,y)$ form.
• Magnitude and direction are recoverable from components via $A=\sqrt{A_x^2+A_y^2}$ and $\theta=\tan^{-1}(A_y/A_x)$ — always double check which quadrant you're in.
• Dot product → scalar, tied to $\cos\theta$. Use it for angles, work, and checking perpendicularity (result = 0).
• Cross product → vector, tied to $\sin\theta$. Use it for torque, angular momentum, and checking parallelism (result = 0).
• Direction of the cross product needs the right-hand rule — magnitude alone doesn't tell you which way it points.
• Sanity-check every answer against the geometry.