Addition of vectors

Walk 3 km east, then 4 km north, and you are not 7 km from where you started — you are 5 km away, along the diagonal shortcut a bird would fly. Ordinary addition (3+4=73+4=7) gets displacement wrong because it ignores direction; vectors need their own arithmetic. That arithmetic is the triangle law and the parallelogram law of vector addition — two equivalent geometric recipes for combining any two vectors, whether they are displacements, velocities, or forces, into a single resultant vector with both a correct magnitude and a correct direction. This article builds that arithmetic from the ground up: how to add two vectors geometrically, the formula that gives the resultant's magnitude and direction without drawing a single triangle, the special cases (same direction, opposite, perpendicular) worth memorising, and how subtraction fits in as "addition of the reverse." Two worked numericals — a right-angle force combination and an equal-vector puzzle — show the formula in action.

The core insight

Two vectors add head-to-tail (triangle law) or as adjacent sides of a parallelogram (parallelogram law) to give a resultant R⃗=A⃗+B⃗\vec R = \vec A + \vec B whose magnitude is R=A2+B2+2ABcos⁡θR=\sqrt{A^2+B^2+2AB\cos\theta} and whose direction follows tan⁡α=Bsin⁡θA+Bcos⁡θ\tan\alpha = \dfrac{B\sin\theta}{A+B\cos\theta}.

The core idea

Triangle law. Draw A⃗\vec A as an arrow. From its head, draw B⃗\vec B (keeping its own length and direction). The vector from the tail of A⃗\vec A to the head of B⃗\vec B is the resultant, R⃗=A⃗+B⃗\vec R = \vec A + \vec B. This is exactly what happens physically when you walk 3 km east and then 4 km north: the two "legs" are A⃗\vec A and B⃗\vec B, and the straight-line trip home is R⃗\vec R.

Parallelogram law. Instead of placing B⃗\vec B at the head of A⃗\vec A, place both vectors tail-to-tail at the same point and complete the parallelogram they form as adjacent sides. The diagonal starting from that common point is the resultant. It is the same answer as the triangle law, just drawn differently — sliding B⃗\vec B to complete the parallelogram is identical to placing it head-to-tail with A⃗\vec A. Both constructions are stated results for this syllabus — their proof is not required, only their statement and use. What is required is the formula for the resultant's size, when the angle between A⃗\vec A and B⃗\vec B is θ\theta:

R=A2+B2+2ABcos⁡θR = \sqrt{A^2 + B^2 + 2AB\cos\theta}

where: AA and BB are the magnitudes of the two vectors, θ\theta is the angle between them (measured with both tails together), and RR is the magnitude of the resultant. The direction of R⃗\vec R, measured as the angle α\alpha it makes with A⃗\vec A, is

tan⁡α=Bsin⁡θA+Bcos⁡θ\tan\alpha = \frac{B\sin\theta}{A + B\cos\theta}

Three special cases are worth memorising because they bound every possible answer. If θ=0°\theta = 0° (same direction), cos⁡θ=1\cos\theta = 1 and R=A+BR = A+B — the maximum possible resultant. If θ=180°\theta = 180° (opposite directions), cos⁡θ=−1\cos\theta = -1 and R=∣A−B∣R = |A-B| — the minimum possible resultant. If θ=90°\theta = 90° (perpendicular), cos⁡θ=0\cos\theta = 0 and R=A2+B2R = \sqrt{A^2+B^2}, the familiar Pythagorean combination. Any computed RR outside the range ∣A−B∣≤R≤A+B|A-B| \le R \le A+B signals an arithmetic slip. Vector addition is commutative (A⃗+B⃗=B⃗+A⃗\vec A+\vec B = \vec B+\vec A) and associative ((A⃗+B⃗)+C⃗=A⃗+(B⃗+C⃗)(\vec A+\vec B)+\vec C=\vec A+(\vec B+\vec C)), so several vectors can be added in any order, in any grouping, one pair at a time by the triangle law — or all at once by the polygon law: place every vector head-to-tail in sequence, and the resultant is the single vector closing the polygon from the very first tail to the very last head.

Subtraction is addition of the negative: A⃗−B⃗=A⃗+(−B⃗)\vec A - \vec B = \vec A + (-\vec B), where −B⃗-\vec B has the same magnitude as B⃗\vec B but points exactly opposite. Drawn from a common origin, A⃗−B⃗\vec A - \vec B runs from the head of B⃗\vec B to the head of A⃗\vec A — reversing the order, B⃗−A⃗\vec B - \vec A, gives a vector of the same length pointing the opposite way.

Figure
Triangle law showing vector A eastward vector B northward from the head of A and the resultant R as the diagonal closing the triangle back to the start
Triangle law — head to tail, then close the triangle A = 3 km east B = 4 km north R = A + B = 5 km start
Derivation AMagnitude and direction of the resultant from the parallelogram law

The goal is to derive R=A2+B2+2ABcos⁡θR=\sqrt{A^2+B^2+2AB\cos\theta} and tan⁡α=Bsin⁡θA+Bcos⁡θ\tan\alpha = \dfrac{B\sin\theta}{A+B\cos\theta} using nothing beyond ordinary right-triangle trigonometry, by dropping a perpendicular from the tip of B⃗\vec B onto the line of A⃗\vec A extended and applying the Pythagorean theorem to the right triangle that appears.

Step 1. Set up the triangle. Place A⃗\vec A along a line, then draw B⃗\vec B from the head of A⃗\vec A at angle θ\theta to A⃗\vec A's direction. The resultant R⃗\vec R runs from the tail of A⃗\vec A to the head of B⃗\vec B. Drop a perpendicular from the head of B⃗\vec B down onto the (extended) line of A⃗\vec A, meeting it at point NN. This perpendicular has length Bsin⁡θB\sin\theta, and the foot NN lies a distance Bcos⁡θB\cos\theta beyond the head of A⃗\vec A.

Step 2. Read off the right triangle. The base of the right triangle, from the tail of A⃗\vec A to NN, has length A+Bcos⁡θA + B\cos\theta (the length of A⃗\vec A plus the extra bit Bcos⁡θB\cos\theta). The height of the same right triangle is Bsin⁡θB\sin\theta. The hypotenuse of this right triangle is exactly RR, since it runs from the tail of A⃗\vec A to the head of B⃗\vec B.

Step 3. Apply the Pythagorean theorem to that right triangle: R2=(A+Bcos⁡θ)2+(Bsin⁡θ)2R^2 = (A+B\cos\theta)^2 + (B\sin\theta)^2. Expand the square: R2=A2+2ABcos⁡θ+B2cos⁡2θ+B2sin⁡2θR^2 = A^2 + 2AB\cos\theta + B^2\cos^2\theta + B^2\sin^2\theta. Since cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1, the two B2B^2 terms collapse to a single B2B^2, giving R2=A2+B2+2ABcos⁡θR^2 = A^2+B^2+2AB\cos\theta, so R=A2+B2+2ABcos⁡θR=\sqrt{A^2+B^2+2AB\cos\theta}.

Step 4. Find the direction. In the same right triangle, the angle α\alpha that R⃗\vec R makes with A⃗\vec A satisfies tan⁡α=oppositeadjacent=Bsin⁡θA+Bcos⁡θ\tan\alpha = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{B\sin\theta}{A+B\cos\theta}, since the opposite side (height) is Bsin⁡θB\sin\theta and the adjacent side (base) is A+Bcos⁡θA+B\cos\theta. Both formulas follow from one right triangle and one trigonometric identity.

Worked example

Two forces of magnitude 6 N and 8 N act on a body at an angle of 90°90° to each other. Find the magnitude and direction of the resultant force, then check what happens if the two forces were equal instead.

Step 1. Identify the values: A=6 NA = 6\ \text{N}, B=8 NB = 8\ \text{N}, θ=90°\theta = 90°, so cos⁡θ=0\cos\theta = 0 and sin⁡θ=1\sin\theta = 1.

Step 2. Magnitude: R=A2+B2+2ABcos⁡θ=62+82+0=36+64=100=10 NR = \sqrt{A^2+B^2+2AB\cos\theta} = \sqrt{6^2+8^2+0} = \sqrt{36+64} = \sqrt{100} = 10\ \text{N}. This is simply the familiar 6-8-10 right triangle — perpendicular vectors always combine by the Pythagorean theorem.

Step 3. Direction: tan⁡α=Bsin⁡θA+Bcos⁡θ=8×16+0=86=1.33\tan\alpha = \dfrac{B\sin\theta}{A+B\cos\theta} = \dfrac{8\times1}{6+0} = \dfrac{8}{6} = 1.33, so α=tan⁡−1(1.33)≈53.1°\alpha = \tan^{-1}(1.33) \approx 53.1° from the 6 N force. The resultant is 10 N, directed at about 53.1 degrees from the 6 N force.

Step 4. Now suppose instead two equal vectors of magnitude 10 units combine to give a resultant that is also 10 units — find the angle between them. Set A=B=R=10A=B=R=10 in the same formula: 10=102+102+2(10)(10)cos⁡θ10 = \sqrt{10^2+10^2+2(10)(10)\cos\theta}, so 100=200+200cos⁡θ100 = 200+200\cos\theta, giving cos⁡θ=−0.5\cos\theta = -0.5 and θ=120°\theta = 120°. Whenever two equal vectors combine to a resultant of the same magnitude, the angle between them is always 120 degrees — a result worth remembering as a shortcut.

Common traps

  • Adding magnitudes directly (e.g. 6 N+8 N=14 N6\ \text{N}+8\ \text{N}=14\ \text{N}) without accounting for the angle — correct only when the two vectors are collinear and in the same direction (θ=0°\theta=0°).
  • Forgetting the bounds ∣A−B∣≤R≤A+B|A-B| \le R \le A+B; any computed resultant outside this range means an arithmetic mistake, not a valid physical answer.
  • Confusing A⃗−B⃗\vec A - \vec B with B⃗−A⃗\vec B - \vec A — same magnitude, exactly opposite direction; order matters.
  • Treating the triangle law and parallelogram law as giving different answers — they are the same construction viewed two ways and must always agree.

Ask the tutor: if the resultant-vector formula or the 120-degree shortcut feel shaky, ask the AI tutor for a few more mixed-angle numericals to build speed.

Why do the triangle law and parallelogram law give the same resultant if they look so different?

They are the same geometric fact drawn two ways. In the parallelogram, slide the vector B⃗\vec B from being an adjacent side (tail at the same point as A⃗\vec A's tail) to sitting head-to-tail with A⃗\vec A instead — its length and direction do not change, so it is still the same B⃗\vec B. The diagonal of the parallelogram and the closing side of the triangle end up being the identical vector, just reached by relabelling which side you started from. Try sketching both for the same two vectors and you will see the resultant arrow lands in exactly the same place.

Is the formula R equals square root of A squared plus B squared always correct?

No — that is only the special case θ=90°\theta = 90°, where cos⁡θ=0\cos\theta = 0 makes the middle term 2ABcos⁡θ2AB\cos\theta vanish from the full formula R=A2+B2+2ABcos⁡θR=\sqrt{A^2+B^2+2AB\cos\theta}. For any other angle you must keep the 2ABcos⁡θ2AB\cos\theta term. Students often apply the Pythagorean shortcut out of habit even when the angle given is not 90°90°, which silently gives the wrong resultant.

How is vector subtraction actually different from vector addition in practice?

Mechanically, subtraction is nothing new: A⃗−B⃗\vec A - \vec B is just A⃗+(−B⃗)\vec A + (-\vec B), so you reverse B⃗\vec B's direction (keeping its magnitude) and then apply the ordinary triangle or parallelogram law to A⃗\vec A and −B⃗-\vec B. The same formula R=A2+B2−2ABcos⁡θR=\sqrt{A^2+B^2-2AB\cos\theta} applies for ∣A⃗−B⃗∣|\vec A - \vec B|, with a minus sign in place of the plus, because reversing B⃗\vec B turns θ\theta into 180°−θ180°-\theta and flips the sign of cos⁡θ\cos\theta in the addition formula.

Before this

Where this leads