Computing and Interpreting Dot Products

Good to see you again. Last time, you learned to express vectors in Cartesian components and to add or subtract them by treating the horizontal and vertical directions separately. The dot product uses those same components, but with a fundamentally different purpose: instead of producing a new vector, it produces one number that measures directional alignment.

In this lesson, you will calculate dot products from components and from magnitudes plus an angle, determine angles between vectors, and interpret positive, zero, and negative results physically. This is an important bridge to work and energy later in the course, and to torque in the next vector lesson.

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A new kind of vector multiplication

Vector addition combines two effects and returns another vector:

$$\vec{A}+\vec{B}=⟨A_x+B_x,A_y+B_y⟩.$$

The dot product, written with a centered dot, combines two vectors and returns a scalar: an ordinary number, possibly with physical units.

For two-dimensional vectors,

$$\vec{A}=⟨A_x,A_y⟩,\vec{B}=⟨B_x,B_y⟩,$$

the dot product is

$$\boxed{\vec{A}\cdot \vec{B}=A_x{B}_x+A_y{B}_y}$$

The procedure is precise:

1. Multiply the matching $x$-components.
2. Multiply the matching $y$-components.
3. Add those two products.

Do not write the result as $⟨A_x{B}_x,A_y{B}_y⟩$. That would still be a vector, whereas a dot product must be a scalar.

Component calculation

Let

$$\vec{p}=⟨5,-2⟩,\vec{q}=⟨-1,4⟩.$$

Then

$$\vec{p}\cdot \vec{q}=(5)(-1)+(-2)(4).$$ $$\boxed{\vec{p}\cdot \vec{q}=-5-8=-13}$$

The negative sign is meaningful; it is not merely an algebraic accident. Shortly, we will see that it tells us these vectors point more oppositely than together.

Notice also that the dot product is commutative:

$$\vec{A}\cdot \vec{B}=\vec{B}\cdot \vec{A}.$$

The matching-component products are the same in either order.

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A vector dotted with itself

The dot product connects directly to the magnitude formula you used in the previous lesson. If

$$\vec{v}=⟨v_x,v_y⟩,$$

then

$$\vec{v}\cdot \vec{v}=v_x^2+v_y^2.$$

But by the Pythagorean theorem,

$$|\vec{v}{|}^2=v_x^2+v_y^2.$$

Therefore,

$$\boxed{\vec{v}\cdot \vec{v}=|\vec{v}{|}^2}$$

For example, with

$$\vec{v}=⟨-3,4⟩,$$

we obtain

$$\vec{v}\cdot \vec{v}=(-3{)}^2+4^2=25.$$

Its magnitude is therefore

$$|\vec{v}|=\sqrt{\vec{v}\cdot \vec{v}}=5.$$

This is a useful consistency check: a vector dotted with itself cannot be negative, because it is the square of its magnitude.

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The geometric meaning: alignment and angle

The component rule is efficient when the components are known. But often in physics, a problem gives the magnitudes of two vectors and the angle between them. The equivalent geometric form of the dot product is

$$\boxed{\vec{A}\cdot \vec{B}=|\vec{A}||\vec{B}|\cos \theta }$$

where $\theta$ is the smallest angle between the two nonzero vectors, from $0$ to $\pi$ radians.

This formula shows that the dot product measures alignment, weighted by the magnitudes of the vectors. The cosine factor identifies how much one vector points along the direction of the other.

To see why this formula makes sense, choose axes so that $\vec{A}$ lies along the positive $x$-axis. Then its components are

$$\vec{A}=⟨|\vec{A}|,0⟩.$$

If $\vec{B}$ makes an angle $\theta$ with $\vec{A}$, its components in those axes are

$$\vec{B}=⟨|\vec{B}|\cos \theta ,|\vec{B}|\sin \theta ⟩.$$

Using the component definition gives

$$\vec{A}\cdot \vec{B}=|\vec{A}|(|\vec{B}|\cos \theta )+0(|\vec{B}|\sin \theta ).$$

Thus,

$$\vec{A}\cdot \vec{B}=|\vec{A}||\vec{B}|\cos \theta .$$

The dot product does not merely ask whether two vectors are long or short. It asks: how much of one vector lies in the direction of the other?

Angle between vectors	Sign of $\cos \theta$	Dot-product interpretation
$0$	Positive maximum	Same direction
Acute angle	Positive	Some shared direction
$\pi /2$	Zero	Perpendicular
Obtuse angle	Negative	More opposite than aligned
$\pi$	Negative minimum	Exactly opposite directions

A dot product of zero is especially important:

$$\boxed{\vec{A}\cdot \vec{B}=0⟺\vec{A}\perp \vec{B}}$$

provided neither vector is the zero vector. The zero vector has no definite direction, so an angle involving it is undefined, although its dot product with every vector is still zero.

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Finding an angle from components

Suppose

$$\vec{A}=⟨2,1⟩,\vec{B}=⟨1,3⟩.$$

First calculate the dot product:

$$\vec{A}\cdot \vec{B}=(2)(1)+(1)(3)=5.$$

Next find each magnitude:

$$|\vec{A}|=\sqrt{2^2+1^2}=\sqrt{5},$$ $$|\vec{B}|=\sqrt{1^2+3^2}=\sqrt{10}.$$

Rearrange the geometric dot-product formula:

$$\cos \theta =\frac{\vec{A}\cdot \vec{B}}{|\vec{A}||\vec{B}|}.$$

Substitute:

$$\cos \theta =\frac{5}{\sqrt{5}\sqrt{10}}=\frac{5}{\sqrt{50}}=\frac{1}{\sqrt{2}}.$$

Therefore,

$$\boxed{\theta =45^{\circ }}$$

The positive dot product predicted an acute angle, and $45^{\circ }$ confirms it.

When using a calculator for a less tidy result, calculate the cosine first and then apply inverse cosine:

$$\theta =\arccos \left(\frac{\vec{A}\cdot \vec{B}}{|\vec{A}||\vec{B}|}\right).$$

Be sure the calculator is in degree mode if a problem requests degrees, or radian mode if it requests radians. Also, the fraction inside inverse cosine must lie between $-1$ and $1$. A value outside that range usually signals an arithmetic error, apart from tiny rounding discrepancies.

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Projection: the quantity the dot product extracts

The expression

$$|\vec{B}|\cos \theta$$

is the signed component of $\vec{B}$ in the direction of $\vec{A}$. It is called the scalar projection of $\vec{B}$ onto $\vec{A}$.

Therefore, the dot product can be read as

$$\vec{A}\cdot \vec{B}=|\vec{A}|\left(\text{component of }\vec{B}\text{ along }\vec{A}\right).$$

This is why a perpendicular component does not affect the dot product. Its projection along the other vector is zero.

For example, imagine moving horizontally while a force has both horizontal and vertical components. The vertical part may matter for other aspects of the situation, but it contributes nothing to the dot product of force with a purely horizontal displacement.

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Physical interpretation: work done by a force

One major physics application is work. For a constant force $\vec{F}$ acting during a displacement $\Delta \vec{r}$,

$$\boxed{W=\vec{F}\cdot \Delta \vec{r}}$$

The unit is

$$\mathrm{N}\cdot \mathrm{m}=\mathrm{J},$$

where $\mathrm{J}$, joule, is the unit of work and energy.

Suppose a force and displacement are

$$\vec{F}=⟨30,40⟩\mathrm{N},$$ $$\Delta \vec{r}=⟨5,0⟩\mathrm{m}.$$

Using components,

$$W=(30\mathrm{N})(5\mathrm{m})+(40\mathrm{N})(0\mathrm{m}).$$ $$\boxed{W=150\mathrm{J}}$$

The $40\mathrm{N}$ vertical part of the force contributes no work because there is no vertical displacement.

We can verify the same result geometrically. The force magnitude is

$$|\vec{F}|=\sqrt{30^2+40^2}\mathrm{N}=50\mathrm{N}.$$

The angle between the force and horizontal displacement has

$$\cos \theta =\frac{30}{50}=\frac{3}{5}.$$

Thus,

$$W=(50\mathrm{N})(5\mathrm{m})\left(\frac{3}{5}\right)=150\mathrm{J}.$$

A positive value means the force has a component in the direction of motion. A negative value means its component is opposite the displacement, as with friction slowing an object. A zero value means the force is perpendicular to the displacement.

This example also illustrates a general units rule: the dot product’s units are the product of the units of the two vectors. For force dotted with displacement, that product is a joule. For a displacement dotted with another displacement, the unit would be ${\mathrm{m}}^2$.

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Choosing the right form and avoiding common errors

In a problem, choose the form that matches the information supplied:

Given information	Efficient method
Both vectors in components	Multiply corresponding components and add
Two magnitudes and their included angle	Use $|\vec{A}||\vec{B}|\cos \theta$
Components but the angle is required	Find the dot product and magnitudes, then use inverse cosine

The main pitfalls are straightforward:

• Returning a vector. A dot product is one scalar, not a pair of components.
• Multiplying without adding. Both matching-component products must be included.
• Dropping negative signs. A negative dot product often has an important interpretation.
• Using the wrong angle. The formula needs the angle between the two vectors, not necessarily either vector’s direction angle measured from the $x$-axis.
• Forgetting units. A physical dot product usually has units, and those units should be multiplied consistently.
• Interpreting zero too broadly. For nonzero vectors, a zero dot product means perpendicular. With a zero vector, it does not determine an angle.

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Wrap-up

The dot product has two equivalent forms:

$$\vec{A}\cdot \vec{B}=A_x{B}_x+A_y{B}_y$$

and

$$\vec{A}\cdot \vec{B}=|\vec{A}||\vec{B}|\cos \theta .$$

The component form is computationally direct; the magnitude-angle form reveals the meaning. A positive result indicates alignment, zero indicates perpendicularity for nonzero vectors, and a negative result indicates opposition. The identity

$$\vec{v}\cdot \vec{v}=|\vec{v}{|}^2$$

also ties the dot product directly to vector magnitude.

Next, you will study the cross product. Unlike the dot product, it produces a vector rather than a scalar, and it captures a different geometric feature: the oriented area and rotational tendency associated with two vectors.