Dot product

Push a lawnmower at an angle and only part of your effort actually moves it forward — the rest is wasted holding the handle down. Physics needs a way to combine two vectors, force and displacement, into the single number that measures "useful effort": that tool is the dot product, also called the scalar product. Given any two vectors inclined at some angle to each other, the dot product multiplies their magnitudes together and scales the result by how well-aligned they are, producing a plain scalar rather than a new vector. It is the natural next step once vectors have been resolved into components, and it immediately pays off in mechanics: work done by a force is defined as exactly this dot product. This article builds the definition, its component form, its geometric meaning, and its most important physical use, work, with fully worked numbers.

The core insight

The dot product a⃗⋅b⃗=abcos⁡θ\vec a\cdot\vec b = ab\cos\theta turns two vectors into the scalar that measures how much they point along each other, and this is precisely how work done by a force is defined.

The core idea

Definition. For two vectors a⃗\vec a and b⃗\vec b with magnitudes aa and bb, inclined at angle θ\theta to each other, the dot product is

a⃗⋅b⃗=abcos⁡θ,\vec a \cdot \vec b = ab\cos\theta,

where: a=∣a⃗∣a=|\vec a|, b=∣b⃗∣b=|\vec b|, and θ\theta is the angle between the two vectors when drawn tail to tail. The result is a scalar — a plain number carrying the product of the two units, never a vector, which is exactly why it is also called the scalar product.

Geometric reading. a⃗⋅b⃗\vec a\cdot\vec b equals (magnitude of a⃗\vec a) times (component of b⃗\vec b along a⃗\vec a) — or, equivalently, (magnitude of b⃗\vec b) times (component of a⃗\vec a along b⃗\vec b). It measures overlap: how much of one vector "lies along" the other, not the vectors' full sizes.

Sign and special cases. Because cos⁡θ\cos\theta ranges from −1-1 to +1+1, the dot product is bounded between −ab-ab and +ab+ab.

  • θ=0°\theta = 0° (parallel, same sense): a⃗⋅b⃗=ab\vec a\cdot\vec b = ab, the maximum.
  • θ=90°\theta = 90° (perpendicular): a⃗⋅b⃗=0\vec a\cdot\vec b = 0 — this single fact is used constantly to spot zero-work forces.
  • θ=180°\theta = 180° (opposite sense): a⃗⋅b⃗=−ab\vec a\cdot\vec b = -ab, the minimum.
  • a⃗⋅a⃗=a2\vec a\cdot\vec a = a^2, since θ=0°\theta = 0° for a vector with itself.
  • Commutative: a⃗⋅b⃗=b⃗⋅a⃗\vec a\cdot\vec b = \vec b\cdot\vec a, because cos⁡θ\cos\theta does not care which vector is named first.

Component form. Resolving each vector along the mutually perpendicular unit vectors ı^,ȷ^,k^\hat\imath,\hat\jmath,\hat k gives a second, purely algebraic route to the same number. Since ı^⋅ı^=ȷ^⋅ȷ^=k^⋅k^=1\hat\imath\cdot\hat\imath=\hat\jmath\cdot\hat\jmath=\hat k\cdot\hat k=1 and all mixed products like ı^⋅ȷ^\hat\imath\cdot\hat\jmath vanish (the unit vectors are perpendicular), multiplying out a⃗=axı^+ayȷ^+azk^\vec a=a_x\hat\imath+a_y\hat\jmath+a_z\hat k against b⃗=bxı^+byȷ^+bzk^\vec b=b_x\hat\imath+b_y\hat\jmath+b_z\hat k leaves only the matching terms:

a⃗⋅b⃗=axbx+ayby+azbz.\vec a \cdot \vec b = a_xb_x + a_yb_y + a_zb_z.

where: (ax,ay,az)(a_x,a_y,a_z) and (bx,by,bz)(b_x,b_y,b_z) are the rectangular components of a⃗\vec a and b⃗\vec b. This form needs no angle at all — useful when vectors are given as components rather than as a magnitude-and-direction pair, and it is also the fastest way to test perpendicularity: two vectors are perpendicular exactly when this sum is zero.

Figure
Two vectors a and b drawn tail to tail at angle theta, with the projection of b onto a shown as a dashed segment whose length equals b cos theta, illustrating that a dot b equals a times that projected length
Dot product as a projection a b theta projection of b along a = b cos theta a dot b = a times b cos theta = a times projected length
Derivation AWork done by a constant force as a dot product

The goal is to show that when a constant force F⃗\vec F acts on a body while it undergoes displacement S⃗\vec S, the work done depends only on the component of the force along the direction of motion, and that this quantity is exactly the dot product F⃗⋅S⃗\vec F\cdot\vec S, so that the perpendicular part of any force does no work whatsoever regardless of its size.

Step 1. Split the force into two perpendicular pieces relative to the displacement direction: a component Fcos⁡θF\cos\theta along S⃗\vec S and a component Fsin⁡θF\sin\theta perpendicular to S⃗\vec S, where θ\theta is the angle between F⃗\vec F and S⃗\vec S. Only the along-displacement piece can push or resist the motion; the perpendicular piece acts sideways to the path and never displaces the body in its own direction.

Step 2. Define work, physically, as (component of force along the displacement) times (magnitude of the displacement), since that is the effective push multiplied by the distance it acts over: W=(Fcos⁡θ) S=FScos⁡θW = (F\cos\theta)\,S = FS\cos\theta.

Step 3. Compare this to the definition of the dot product, F⃗⋅S⃗=FScos⁡θ\vec F\cdot\vec S = FS\cos\theta with F=∣F⃗∣F=|\vec F| and S=∣S⃗∣S=|\vec S|. The two expressions are identical term for term, so W=F⃗⋅S⃗W = \vec F\cdot\vec S — work is precisely the dot product of force and displacement, no extra factor needed.

Step 4. Read off the two limiting cases that examiners favour: at θ=0°\theta=0°, W=FSW=FS, the full possible work is done because the force points exactly along the motion; at θ=90°\theta=90°, W=0W=0, because a force entirely sideways to the motion (like the normal reaction on a body sliding along a horizontal floor) does no work at all, however large it is.

Worked example

A force F⃗=10 N\vec F = 10\ \text{N} acts at 60°60° to a displacement of magnitude 5 m5\ \text{m}. A second question then asks for the dot product of a⃗=3ı^+4ȷ^−2k^\vec a = 3\hat\imath+4\hat\jmath-2\hat k and b⃗=−1ı^+2ȷ^+5k^\vec b=-1\hat\imath+2\hat\jmath+5\hat k, and whether the angle between them is acute or obtuse.

Step 1. Work from the angle form: W=F⃗⋅S⃗=FScos⁡θ=(10)(5)cos⁡60°=50×0.5=25 JW = \vec F\cdot\vec S = FS\cos\theta = (10)(5)\cos60° = 50\times0.5 = 25\ \text{J}. Only the component of force along the motion, 10cos⁡60°=5 N10\cos60°=5\ \text{N}, does any work; the perpendicular part contributes nothing, which is why the answer is half of the maximum possible 10×5=50 J10\times5=50\ \text{J}.

Step 2. Dot product from components: a⃗⋅b⃗=(3)(−1)+(4)(2)+(−2)(5)=−3+8−10=−5\vec a\cdot\vec b = (3)(-1)+(4)(2)+(-2)(5) = -3+8-10=-5.

Step 3. Interpret the sign. Since a⃗⋅b⃗=abcos⁡θ\vec a\cdot\vec b = ab\cos\theta with a,b>0a,b>0 always, the sign of the dot product is the sign of cos⁡θ\cos\theta. Here a⃗⋅b⃗=−5<0\vec a\cdot\vec b=-5<0, so cos⁡θ<0\cos\theta<0 and θ\theta is obtuse — between 90°90° and 180°180° — without ever computing θ\theta itself.

Answer: the force does 25 J25\ \text{J} of work, and a⃗⋅b⃗=−5\vec a\cdot\vec b=-5, telling us the angle between a⃗\vec a and b⃗\vec b is obtuse.

Common traps

  • Treating the dot product as if it produces a vector — it is always a scalar, with no direction of its own.
  • Forgetting perpendicular vectors give zero: a force acting exactly at 90°90° to the displacement (normal reaction, gravity on a horizontal path) does zero work, no matter its magnitude.
  • Swapping cosine and sine: abcos⁡θab\cos\theta belongs to the dot product; absin⁡θab\sin\theta belongs to the cross product, a different tool entirely.
  • Assuming the dot product can exceed abab in size — it is always bounded between −ab-ab and +ab+ab, reaching those extremes only at θ=0°\theta=0° and θ=180°\theta=180°.

Ask the tutor: if work-done-at-an-angle numericals or component dot products still feel shaky, ask the AI tutor for a few more practice problems.

Why does a perpendicular force do zero work if it clearly holds the object up?

The normal reaction (or any perpendicular force) certainly matters for balance, but "work" in physics is defined narrowly as force times displacement in the direction of the force. If the object only moves sideways while the force points straight up, the up-force never actually pushes the object any distance along its own direction, so cos⁡90°=0\cos90°=0 makes the dot product, and hence the work, exactly zero. The force can support the weight without ever doing work on the motion.

How is the component formula a_x b_x + a_y b_y + a_z b_z the same thing as ab cos theta?

Both are the same number reached two different ways. Write each vector in unit-vector form and multiply term by term; because ı^,ȷ^,k^\hat\imath,\hat\jmath,\hat k are mutually perpendicular unit vectors, every cross term like ı^⋅ȷ^\hat\imath\cdot\hat\jmath vanishes and only the matching axbxa_xb_x, aybya_yb_y, azbza_zb_z terms survive. That algebra is exactly equivalent to abcos⁡θab\cos\theta — the component route just avoids ever needing to know θ\theta explicitly, which is handy when vectors are given as coordinates rather than as magnitude-and-direction.

Is the dot product only useful for work, or does it show up elsewhere too?

Within this syllabus, work done by a constant force is the headline application, and it is the one you should be fluent in for numericals. But the dot product itself is a general tool for measuring alignment between any two vectors — its sign tells you instantly whether an angle is acute, right, or obtuse, which is a quick geometry check worth remembering even outside mechanics problems.

Before this

Where this leads