Physics › 📐 Foundations

Vectors & Components

Some quantities need a number and a direction to mean anything — here's how to break them apart and put them back together.

High school
💡
The big idea: A vector packs two pieces of information into one arrow: how much, and which way. The real power comes from splitting that arrow into a horizontal piece and a vertical piece that don't interfere with each other — you can add, subtract, and reason about each direction on its own, then recombine them whenever you need the full picture back.
🎯 By the end, you'll be able to
  • distinguish scalar quantities from vector quantities and give clear real-world examples of each
  • resolve any vector into perpendicular x- and y-components using sine and cosine
  • add two or more vectors either by drawing them head-to-tail or by adding their components algebraically
  • calculate the magnitude and direction of a resultant vector from its components, correctly placing the angle in the right quadrant
📎 You should already know
  • Basic trigonometry (sine, cosine, tangent of a right triangle)
  • The Pythagorean theorem
  • Plotting points on an x-y plane

Not everything in physics is just a number

Tell a friend "drive 10 miles" and they'll rightly ask, "which way?" Tell them "drive 10 miles north" and now they can actually get somewhere. That's the whole difference between a scalar and a vector. Mass, temperature, time, and speed are scalars — a single number tells the whole story. Displacement, velocity, force, and acceleration are vectors — you need a size and a direction, or the description is incomplete.

Once a quantity has a direction, arrows become the natural language for it: the length of the arrow shows magnitude, and the way it points shows direction.

🔑 The core trick: split it, then add it back

Any vector, no matter which way it points, can be replaced by two perpendicular vectors — one purely horizontal, one purely vertical — that together have the exact same effect as the original. These are its components. The payoff is huge: horizontal pieces only ever interact with other horizontal pieces, and vertical pieces only with other vertical pieces. That turns messy diagonal-arrow problems into simple side-by-side addition.

\[ v_x = v\cos\theta \qquad v_y = v\sin\theta \]
θ is measured from the positive x-axis; v is the vector's magnitude (its length).
\[ |v| = \sqrt{v_x^2 + v_y^2} \qquad \theta = \tan^{-1}\!\left(\frac{v_y}{v_x}\right) \]
Going the other way: rebuild the magnitude and direction once you know the components.
🎮 Interactive: Build and Add Vectors LIVE
Drag vectors around to place them head-to-tail, watch their x- and y-components light up separately, and see the resultant arrow form in real time.

Two roads to the same answer

There are two equivalent ways to add vectors, and it's worth knowing both. Head-to-tail (graphical): draw the first vector, then start the second vector's tail exactly where the first one's head ends. The resultant is the single arrow drawn from the very first tail to the very last head — order doesn't matter, you'll land in the same place either way.

By components (algebraic): resolve every vector into its \(x\)- and \(y\)-parts, add all the \(x\)-parts together, add all the \(y\)-parts together, and those two sums are the components of the resultant. This is the method you'll use for almost every real calculation, because it works no matter how many vectors you're combining or what angles they're at.

📝 Worked example: A hiking trail sign says: walk 3 km east, then 4 km north. Find the magnitude and direction of your total displacement.
  1. Set up components for each leg: East 3 km is purely horizontal, so \(A_x = 3\) km, \(A_y = 0\) km.
  2. North 4 km is purely vertical, so \(B_x = 0\) km, \(B_y = 4\) km.
  3. Add matching components: \(R_x = A_x + B_x = 3 + 0 = 3\) km, and \(R_y = A_y + B_y = 0 + 4 = 4\) km.
  4. Find the magnitude: \(|R| = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5\) km.
  5. Find the direction: \(\theta = \tan^{-1}(4/3) \approx 53.1^\circ\) north of east.
✓ The total displacement is 5 km, directed 53.1° north of east — noticeably shorter than the 7 km actually walked, because straight-line distance cuts the corner.
📝 Worked example: A boat's engine pushes it at 20 m/s in a direction 60° north of east. A current adds an extra 5 m/s due north. Find the boat's resultant speed and direction.
  1. Resolve the engine's velocity into components: \(v_x = 20\cos(60^\circ) = 20(0.5) = 10\) m/s, \(v_y = 20\sin(60^\circ) = 20(0.866) \approx 17.3\) m/s.
  2. The current is purely north, so its components are \(0\) m/s east and \(5\) m/s north.
  3. Add components: \(R_x = 10 + 0 = 10\) m/s, \(R_y = 17.3 + 5 = 22.3\) m/s.
  4. Magnitude: \(|R| = \sqrt{10^2 + 22.3^2} = \sqrt{100 + 497.3} \approx \sqrt{597.3} \approx 24.4\) m/s.
  5. Direction: \(\theta = \tan^{-1}(22.3/10) \approx 65.8^\circ\) north of east.
✓ The boat actually moves at about 24.4 m/s, roughly 65.8° north of east — faster than the engine alone, and pushed further off its heading by the current.
⚠️ The arctan trap: your calculator doesn't know which quadrant you're in

\(\tan^{-1}\) only ever returns an angle between \(-90^\circ\) and \(90^\circ\), because it has no way of knowing whether both components were positive, both negative, or one of each. Before trusting the number your calculator gives you, look at the signs of \(v_x\) and \(v_y\) and sketch a rough picture: positive-positive lands in the upper right, negative-positive in the upper left, and so on. A vector with \(v_x = -3\) and \(v_y = 4\) is not at the same angle as one with \(v_x = 3\) and \(v_y = -4\), even though \(\tan^{-1}(4/-3)\) and \(\tan^{-1}(-4/3)\) can spit out deceptively similar-looking results.

Check your understanding

1. Which of the following is a vector quantity?
Velocity needs both a speed and a direction to be fully described, which makes it a vector. Mass, temperature, and energy are complete with just a single number — they're scalars.
2. A force of 10 N points at 37° above the horizontal. Using cos(37°) ≈ 0.8 and sin(37°) ≈ 0.6, what is its horizontal component?
The horizontal component is \(F_x = F\cos\theta = 10 \times 0.8 = 8\) N. The 6 N figure is actually the vertical component, \(F_y = 10 \times 0.6\).
3. Two vectors have equal magnitude but point in exactly opposite directions. What is the magnitude of their sum?
Opposite directions mean their components are equal in size but opposite in sign, so when you add matching components they cancel completely, leaving a resultant of zero.
4. A vector has components \(v_x = 3\) and \(v_y = 4\) (in meters). What are its magnitude and direction above the x-axis?
Magnitude: \(\sqrt{3^2+4^2} = \sqrt{25} = 5\) m. Direction: \(\tan^{-1}(4/3) \approx 53.1^\circ\) — note the angle comes from \(v_y/v_x\), not the other way around, which is the mistake that leads to the tempting-but-wrong 37° answer.
✅ Key takeaways
  • Scalars have magnitude only (mass, time, speed); vectors have both magnitude and direction (displacement, velocity, force).
  • Any vector resolves into perpendicular components with v_x = v cos θ and v_y = v sin θ, measured from the x-axis.
  • Vectors add either by drawing them head-to-tail or, more reliably, by adding their matching x- and y-components separately.
  • Recover the resultant's magnitude with the Pythagorean theorem and its direction with arctan(v_y/v_x) — but always check the component signs to place the angle in the correct quadrant.