Scalar (Dot) Product

You are at a railway station, dragging a heavy suitcase across the platform to catch a train. The handle is at an angle — you are pulling upward and forward at the same time. Common sense tells you that only the forward part of your pull actually moves the suitcase forward. The upward part lifts against gravity but does not move the suitcase along the platform. So if someone asked "how much of your force is actually doing the job of moving the suitcase?", the answer is not the full force — it is the piece of the force that points in the direction the suitcase is moving.

That "piece of the force in the direction of motion" is exactly what the dot product calculates. It takes two vectors — your force and the displacement of the suitcase — and returns a single number that captures how much one vector acts along the other. That number is not a vector. It has no direction. It is a pure scalar, which is why this operation is called the scalar product.

The dot product shows up everywhere in physics: work, power, flux, the angle between two directions, testing whether two vectors are perpendicular. Every time you need to ask "how much of this vector points along that vector?", the answer is a dot product.

The geometry: projection

Before writing any formula, look at what the dot product is doing geometrically.

Take two vectors \vec{A} and \vec{B} placed tail-to-tail, with an angle \theta between them. Drop a perpendicular from the tip of \vec{A} onto the line of \vec{B}. The foot of that perpendicular marks a length along \vec{B} — this length is called the projection of \vec{A} onto \vec{B}, and it equals |\vec{A}|\cos\theta.

Think of it as a shadow. If you shine a light straight down (perpendicular to \vec{B}), the shadow of \vec{A} on the line of \vec{B} has length |\vec{A}|\cos\theta. When \theta = 0 (the vectors point in the same direction), the shadow is the full length of \vec{A}. When \theta = 90°, the shadow has zero length — \vec{A} is perpendicular to \vec{B} and contributes nothing along it.

The dot product takes this projection and multiplies it by the magnitude of \vec{B}:

Why multiply by |\vec{B}|: The projection |\vec{A}|\cos\theta tells you how much of \vec{A} points along \vec{B}, but it does not know how long \vec{B} is. Multiplying by |\vec{B}| scales the result so that both magnitudes matter symmetrically — the dot product is just as much "the projection of \vec{B} onto \vec{A}, times |\vec{A}|" as it is the other way around.

Reading the definition. |\vec{A}| is the magnitude (length) of \vec{A}. |\vec{B}| is the magnitude of \vec{B}. \cos\theta encodes the alignment between them: it equals 1 when they are parallel (\theta = 0°), 0 when they are perpendicular (\theta = 90°), and -1 when they are anti-parallel (\theta = 180°). So the dot product is largest when the vectors are aligned, zero when they are at right angles, and negative when they point in roughly opposite directions.

Notice how the \cos\theta factor does all the work of extracting "how much of one vector points along the other." The magnitudes just scale the answer.

The dot product in component form

The geometric definition is elegant, but in practice you rarely know the angle \theta between two vectors. You usually know their components. So you need a way to compute the dot product from components alone — and deriving this form teaches you something about unit vectors along the way.

Write the two vectors in component form using the unit vectors \hat{i}, \hat{j}, \hat{k}:

To find \vec{A} \cdot \vec{B}, use the distributive property (which you will prove in the next section) and expand:

Why: treat the dot product like multiplication — each term in the first bracket dots with each term in the second. This gives nine terms.

Now apply the geometric definition to the unit vector dot products. Each unit vector has magnitude 1, and the angles between them are known:

• \hat{i}\cdot\hat{i} = |\hat{i}||\hat{i}|\cos 0° = 1 \times 1 \times 1 = 1. Similarly, \hat{j}\cdot\hat{j} = 1 and \hat{k}\cdot\hat{k} = 1.

Why: any unit vector dotted with itself gives \cos 0° = 1. The vector is perfectly aligned with itself.

• \hat{i}\cdot\hat{j} = |\hat{i}||\hat{j}|\cos 90° = 1 \times 1 \times 0 = 0. Similarly, \hat{j}\cdot\hat{k} = 0 and \hat{i}\cdot\hat{k} = 0.

Why: the coordinate axes are perpendicular to each other, so the angle between any two different unit vectors is 90°, and \cos 90° = 0.

Six of the nine terms vanish. The three survivors are the terms where the same unit vector appears on both sides:

Why this is powerful: you no longer need to know the angle \theta. Given the components of two vectors, you multiply corresponding components and add. Three multiplications and two additions — that is the entire calculation.

This is the component form of the dot product. It is equivalent to the geometric form; they give the same number. The component form is what you will use in almost every calculation.

Finding the angle between two vectors

Combining the two forms gives you a tool for finding angles. Since both forms give the same result:

Solve for \cos\theta:

Why: you used the component form to compute the numerator and the magnitude formula to compute the denominator. The angle \theta drops out from the geometric form. This is the standard method for finding the angle between two vectors when you know their components.

This formula is the reason the dot product matters in geometry. Given any two vectors in 2D or 3D, you can find the angle between them without drawing anything — just plug in the components.

Properties of the dot product

1. Commutative: \vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}

From the definition: |\vec{A}||\vec{B}|\cos\theta = |\vec{B}||\vec{A}|\cos\theta. Multiplication of real numbers is commutative, so the order does not matter.

Why this matters: unlike the cross product (which flips sign when you swap the order), the dot product does not care which vector comes first.

2. Distributive: \vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C}

This is the property that let you expand the nine-term product earlier. To see why it holds, think in terms of projections. The projection of (\vec{B} + \vec{C}) onto \vec{A} is the sum of the individual projections — adding the vectors first and then projecting gives the same result as projecting each one separately and then adding. The distributive law is the algebraic statement of this geometric fact.

Why this matters: it lets you expand dot products involving sums of vectors, just like expanding brackets in ordinary algebra. Without this property, the component form derivation above would not work.

3. Scalar multiplication: (c\vec{A}) \cdot \vec{B} = c(\vec{A} \cdot \vec{B})

If you scale a vector by a constant c, its magnitude gets multiplied by |c| and (if c is negative) its direction reverses. The dot product scales linearly:

When c > 0, \theta' = \theta and this equals c|\vec{A}||\vec{B}|\cos\theta = c(\vec{A}\cdot\vec{B}). When c < 0, the direction of \vec{A} reverses so \theta' = 180° - \theta, and \cos(180° - \theta) = -\cos\theta. This gives |c||\vec{A}||\vec{B}|(-\cos\theta) = c(\vec{A}\cdot\vec{B}), since c = -|c|.

4. Dot product with itself: \vec{A} \cdot \vec{A} = |\vec{A}|^2

Set \vec{B} = \vec{A} in the definition. The angle between a vector and itself is 0°, so \cos 0° = 1:

Why this is useful: it connects the dot product to the magnitude. You can always find the length of a vector by taking |\vec{A}| = \sqrt{\vec{A}\cdot\vec{A}}. In component form: |\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2}.

5. Zero when perpendicular: \vec{A} \cdot \vec{B} = 0 \iff \vec{A} \perp \vec{B}

When \theta = 90°, \cos 90° = 0, so \vec{A} \cdot \vec{B} = 0. And conversely, if \vec{A} \cdot \vec{B} = 0 (and neither vector is the zero vector), then \cos\theta = 0, which means \theta = 90°. The vectors are perpendicular.

This is the perpendicularity test. To check whether two vectors are at right angles, compute their dot product. If it is zero, they are perpendicular. No trigonometry, no protractor, no drawing — just arithmetic.

Summary table

Work done by a force — the dot product in action

The single most important application of the dot product in mechanics is the definition of work.

Imagine you are pushing a heavy almirah across a room. You push with a force \vec{F} and the almirah slides through a displacement \vec{d}. If your force is exactly in the direction of motion, every bit of your effort goes into moving the almirah. But if your force is at an angle — say you are pushing partly downward — only the component of force along the direction of motion does work. The component perpendicular to the motion just presses the almirah into the floor.

The work done by the force is:

Why \cos\theta: the factor F\cos\theta is the component of \vec{F} in the direction of \vec{d}. Only this component contributes to moving the object. The perpendicular component F\sin\theta pushes the object into the floor (or lifts it off the floor) but does not displace it forward.

Three special cases make this concrete:

• \theta = 0° (force in the direction of motion): W = Fd. All of your effort goes into moving the object. Example: dragging a boat along a canal with the rope horizontal.

• \theta = 90° (force perpendicular to motion): W = 0. You exert a force but do no work. Example: carrying a heavy bag while walking on a level road — gravity pulls the bag down, you walk forward, and gravity does zero work because the force (downward) is perpendicular to the displacement (horizontal).

• \theta = 180° (force opposite to motion): W = -Fd. The force opposes the motion. Example: friction on the almirah as you push it — friction points backward while the almirah moves forward, so friction does negative work (removes energy from the almirah).

Worked examples

Common confusions

• "The dot product gives a vector." No — the dot product of two vectors is always a scalar (a plain number). That is the whole point of calling it the "scalar product." If you want an operation that gives a vector from two vectors, that is the cross product.

• "\vec{A}\cdot\vec{B} = AB." Only when \theta = 0° (the vectors are parallel and in the same direction). In general, \vec{A}\cdot\vec{B} = AB\cos\theta. Forgetting the \cos\theta is the most common error in work calculations — it makes you overestimate the work done by a force that is not in the direction of motion.

• "A negative dot product means the vectors are perpendicular." No — a negative dot product means the angle between the vectors is greater than 90° (they point in roughly opposite directions). Perpendicular vectors have a dot product of exactly zero, not a negative number.

• "You can dot a vector with a scalar." The dot product is defined only between two vectors. Writing \vec{A} \cdot 5 is meaningless. If you want to scale a vector by a number, that is ordinary scalar multiplication: 5\vec{A}.

• "Work is always positive." Work can be negative. When the force opposes the displacement (\theta > 90°), \cos\theta < 0 and the work is negative. Friction always does negative work — it opposes motion, so the angle between friction and displacement is 180°, and \cos 180° = -1.

• "If the dot product is zero, one of the vectors must be zero." Not necessarily. Two nonzero vectors can have a zero dot product — they just need to be perpendicular. The vectors 3\hat{i} + 4\hat{j} and 4\hat{i} - 3\hat{j} are both nonzero, but their dot product is 3(4) + 4(-3) = 12 - 12 = 0. They are at right angles.

Where this leads next

• Vector (Cross) Product — the other way to multiply two vectors, giving a vector result perpendicular to both. Where the dot product measures alignment, the cross product measures "how much the vectors span a parallelogram."
• Work and Energy — the full treatment of work done by constant and variable forces, building on W = \vec{F}\cdot\vec{d}.
• Power — instantaneous power is P = \vec{F}\cdot\vec{v}, another dot product — force dotted with velocity.
• Vectors: Introduction and Notation — if the component notation in this article felt unfamiliar, start here.
• Vector Addition and Subtraction — the operations you need before you can use the distributive property of the dot product.