Addition of vectors

Walk three blocks east, then four blocks north, and your net displacement is a single straight arrow from start to finish, not the sum of the two distances you actually covered. That arrow — five blocks, at an angle — is what vector addition captures: two quantities that have both size and direction combine into a third quantity that is genuinely new, not just "add the numbers." The same idea handles two forces pulling on a hook, two velocities of wind and aircraft, or two displacements in a 3D geometry problem. This section builds the two equivalent geometric pictures for adding vectors — the triangle law and the parallelogram law — and the algebraic component form that turns either picture into fast, exact arithmetic. Component addition, in particular, is the workhorse you will reuse constantly for the section formula, the dot and cross products, and 3D-geometry problems later in the Vector Algebra chapter.

The core insight

Vectors add head-to-tail (triangle law) or as parallelogram diagonals (same result, different picture), and in component form a⃗+b⃗\vec a+\vec b is found simply by adding matching components.

The core idea

Triangle law. Place the tail of b⃗\vec b at the head of a⃗\vec a; the sum a⃗+b⃗\vec a+\vec b is the third side of the triangle, drawn from the free tail of a⃗\vec a to the free head of b⃗\vec b. In symbols, if AB→=a⃗\overrightarrow{AB}=\vec a and BC→=b⃗\overrightarrow{BC}=\vec b,

AB→+BC→=AC→,\overrightarrow{AB}+\overrightarrow{BC} = \overrightarrow{AC},

where: AA, BB, CC are points, and AC→\overrightarrow{AC} is the single vector representing the combined effect of the two displacements in sequence.

Parallelogram law. Draw a⃗\vec a and b⃗\vec b as two adjacent sides of a parallelogram from the same starting point; the sum a⃗+b⃗\vec a+\vec b is the diagonal through that point. This is not a second rule but the same triangle seen twice — the parallelogram is two copies of one triangle glued back to back along a shared diagonal, so both laws always agree.

Properties of addition. For any vectors a⃗,b⃗,c⃗\vec a,\vec b,\vec c:

a⃗+b⃗=b⃗+a⃗,(a⃗+b⃗)+c⃗=a⃗+(b⃗+c⃗),\vec a+\vec b = \vec b+\vec a, \qquad (\vec a+\vec b)+\vec c = \vec a+(\vec b+\vec c),
a⃗+0⃗=a⃗,a⃗+(−a⃗)=0⃗,\vec a+\vec 0=\vec a, \qquad \vec a+(-\vec a)=\vec 0,

where: 0⃗\vec 0 is the zero vector (no magnitude, no fixed direction) and −a⃗-\vec a is a⃗\vec a reversed in direction with the same magnitude. Commutativity and associativity together mean that a chain of several vectors can be added head to tail in any order, and the sum is always the single arrow from the very first tail to the very last head.

Component form. With a⃗=a1i^+a2j^+a3k^\vec a=a_1\hat i+a_2\hat j+a_3\hat k and b⃗=b1i^+b2j^+b3k^\vec b=b_1\hat i+b_2\hat j+b_3\hat k,

a⃗+b⃗=(a1+b1)i^+(a2+b2)j^+(a3+b3)k^,\vec a+\vec b=(a_1+b_1)\hat i+(a_2+b_2)\hat j+(a_3+b_3)\hat k,

where: i^,j^,k^\hat i,\hat j,\hat k are unit vectors along the coordinate axes and a1,a2,a3a_1,a_2,a_3 are a⃗\vec a's components along them. Subtraction is the same idea applied to a⃗+(−b⃗)\vec a+(-\vec b), so components simply subtract: a⃗−b⃗=(a1−b1)i^+(a2−b2)j^+(a3−b3)k^\vec a-\vec b=(a_1-b_1)\hat i+(a_2-b_2)\hat j+(a_3-b_3)\hat k. The triangle law also generalises to any number of vectors: to add a⃗+b⃗+c⃗\vec a+\vec b+\vec c, place the vectors head to tail in sequence and join the very first tail to the very last head. Because addition is both commutative and associative, the order in which the vectors are placed does not change the final resultant — a fact worth remembering when a problem gives several forces or displacements at once rather than just two. The picture below shows why both laws must give the same diagonal: the same two vectors, drawn once as a triangle and once as a parallelogram, land on an identical resultant.

Figure
Two vectors combined by the triangle law and by the parallelogram law giving the same resultant arrow
Triangle law and parallelogram law give the same sum triangle law a b a+b tail of b meets head of a parallelogram law a b a+b diagonal from common vertex

Derivation

Derivation AAdding vectors from PQ and QR to get PR

The goal is to find PR→\overrightarrow{PR} when PQ→=2i^+3j^−k^\overrightarrow{PQ}=2\hat i+3\hat j-\hat k and QR→=−i^+4j^+2k^\overrightarrow{QR}=-\hat i+4\hat j+2\hat k are given as two displacements that share the common point QQ, demonstrating that the triangle law reduces to plain component addition once the vectors are already tail-to-head.

Step 1. Recognise the shared point. QQ is the head of PQ→\overrightarrow{PQ} and the tail of QR→\overrightarrow{QR}, so the two vectors are already positioned exactly as the triangle law requires: no translation is needed before adding.

Step 2. Apply the triangle law. Since the vectors are placed head to tail through QQ, their sum is the third side of the triangle, running from PP directly to RR:

PR→=PQ→+QR→.\overrightarrow{PR}=\overrightarrow{PQ}+\overrightarrow{QR}.

Step 3. Add components. Matching i^,j^,k^\hat i,\hat j,\hat k parts of PQ→=2i^+3j^−k^\overrightarrow{PQ}=2\hat i+3\hat j-\hat k and QR→=−i^+4j^+2k^\overrightarrow{QR}=-\hat i+4\hat j+2\hat k term by term,

PR→=(2+(−1))i^+(3+4)j^+(−1+2)k^=i^+7j^+k^.\overrightarrow{PR}=(2+(-1))\hat i+(3+4)\hat j+(-1+2)\hat k=\hat i+7\hat j+\hat k.

Step 4. State and, optionally, check the result. So PR→=i^+7j^+k^\overrightarrow{PR}=\hat i+7\hat j+\hat k; its magnitude is ∣PR→∣=12+72+12=51|\overrightarrow{PR}|=\sqrt{1^2+7^2+1^2}=\sqrt{51}, a useful sanity figure since it must be less than ∣PQ→∣+∣QR→∣|\overrightarrow{PQ}|+|\overrightarrow{QR}| by the triangle inequality.

Worked example

Two vectors a⃗=5i^−2j^+k^\vec a=5\hat i-2\hat j+\hat k and b⃗=−3i^+j^+4k^\vec b=-3\hat i+\hat j+4\hat k act along adjacent sides of a parallelogram from a common vertex. Find the vector along the diagonal through that vertex, and the vector along the other diagonal.

Step 1. Diagonal through the common vertex. By the parallelogram law, this diagonal is a⃗+b⃗\vec a+\vec b, found by adding components:

a⃗+b⃗=(5−3)i^+(−2+1)j^+(1+4)k^=2i^−j^+5k^.\vec a+\vec b=(5-3)\hat i+(-2+1)\hat j+(1+4)\hat k=2\hat i-\hat j+5\hat k.

Step 2. The other diagonal. This diagonal connects the tips of a⃗\vec a and b⃗\vec b, and is given by the difference b⃗−a⃗\vec b-\vec a:

b⃗−a⃗=(−3−5)i^+(1−(−2))j^+(4−1)k^=−8i^+3j^+3k^.\vec b-\vec a=(-3-5)\hat i+(1-(-2))\hat j+(4-1)\hat k=-8\hat i+3\hat j+3\hat k.

The diagonal through the common vertex is 2i^−j^+5k^2\hat i-\hat j+5\hat k, and the other diagonal is −8i^+3j^+3k^-8\hat i+3\hat j+3\hat k (or its negative, a⃗−b⃗\vec a-\vec b, depending on which endpoint is taken as the start).

Common traps

  • Adding magnitudes instead of vectors. ∣a⃗+b⃗∣≠∣a⃗∣+∣b⃗∣|\vec a+\vec b|\neq|\vec a|+|\vec b| in general — equality holds only when a⃗\vec a and b⃗\vec b point the same way; otherwise ∣a⃗+b⃗∣≤∣a⃗∣+∣b⃗∣|\vec a+\vec b|\le|\vec a|+|\vec b|.
  • Misplacing the tail of the second vector. The triangle law reads off directly as AC→\overrightarrow{AC} only when BC→\overrightarrow{BC}'s tail sits exactly at AB→\overrightarrow{AB}'s head; vectors given from different starting points must first be translated (valid, since equal vectors are position-independent).
  • Mixing up the two diagonals of a parallelogram. The diagonal from the common initial point is the sum a⃗+b⃗\vec a+\vec b; the diagonal joining the two free endpoints is the difference b⃗−a⃗\vec b-\vec a.
  • Treating subtraction as commutative. a⃗−b⃗=−(b⃗−a⃗)\vec a-\vec b=-(\vec b-\vec a), not equal to b⃗−a⃗\vec b-\vec a; only addition commutes.

Ask the tutor: want a walkthrough showing exactly why the parallelogram's two diagonals are a⃗+b⃗\vec a+\vec b and a⃗−b⃗\vec a-\vec b, or practice translating vectors before applying the triangle law — just ask.

Why do the triangle law and the parallelogram law always give the same answer?

Because they are the same construction viewed two ways. In the parallelogram, the diagonal from the common vertex splits the shape into two congruent triangles, and each triangle is exactly the triangle-law picture: one side is a⃗\vec a, the next side (translated so its tail sits at a⃗\vec a's head) is b⃗\vec b, and the diagonal is a⃗+b⃗\vec a+\vec b. Use the triangle law when vectors are naturally given tail-to-head, like a sequence of displacements, and the parallelogram law when two vectors act together from one point, like two forces on a hook — the arithmetic answer is identical either way.

Do I need to translate a vector before adding it, and is that allowed?

Yes, and yes it is allowed. Two vectors are considered equal as long as they have the same magnitude and direction, regardless of where they are drawn — a vector has no fixed "location," only size and direction. So if PQ→\overrightarrow{PQ} and RS→\overrightarrow{RS} are not naturally head-to-tail, you can slide one of them (keeping its length and direction fixed) until its tail meets the other's head, then apply the triangle law. In component form this translation step is invisible: you simply add the i^,j^,k^\hat i,\hat j,\hat k parts directly, without needing to draw anything.

What is the fastest way to add vectors given in component form, without drawing a diagram?

Just add matching components: if a⃗=a1i^+a2j^+a3k^\vec a=a_1\hat i+a_2\hat j+a_3\hat k and b⃗=b1i^+b2j^+b3k^\vec b=b_1\hat i+b_2\hat j+b_3\hat k, then a⃗+b⃗=(a1+b1)i^+(a2+b2)j^+(a3+b3)k^\vec a+\vec b=(a_1+b_1)\hat i+(a_2+b_2)\hat j+(a_3+b_3)\hat k. No diagram is required because the triangle and parallelogram laws are already baked into this formula — component addition is simply the algebraic shortcut for either geometric picture. Subtraction works the same way, term by term, and this is the method you should default to whenever vectors are given in i^,j^,k^\hat i,\hat j,\hat k form rather than as a diagram.

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Where this leads