Dot Product vs Cross Product: What Is the Difference?

The dot product asks how much of one vector points along the other and returns a scalar, A B cosine theta. The cross product asks how much points perpendicular and returns a vector of magnitude A B sine theta. That is why work is a scalar and torque is a vector.

AP Physics: Unit 3 (topics 3.2 Work, 5.3 Torque, 8.5 Electric Flux, 12.2 Magnetism and Moving Charges). Both products are AP Physics C content only. The AP Physics C: Mechanics CED defines the dot product at EK 3.2.A.3.i, inside Topic 3.2, Work, as resulting in a scalar quantity of magnitude AB cosine theta, alongside the general work definition at EK 3.2.A.3 as the integral of F dotted with dr over the path. It defines the cross-product at EK 5.3.B.2.i, inside Topic 5.3, Torque, as resulting in a vector quantity of magnitude AB sine theta, with EK 5.3.B.2.ii giving the direction as perpendicular to both vectors and normal to the plane they define, and EK 5.3.B.2.iii giving the right-hand rule. Suggested skills for Topic 5.3 in C: Mechanics are 1.A, 2.B, 2.C and 3.C. The AP Physics C: Electricity and Magnetism CED uses the dot product at EK 8.5.A.2 and 8.5.A.2.ii for electric flux and at EK 13.1.A.1.ii for magnetic flux, and the cross-product at EK 12.2.B.1 for the magnetic force on a moving charge, with the relevant equation F_B = q times v cross B. Suggested skills for Topic 8.5 are 1.A, 2.A, 2.C and 3.B, and for Topic 12.2 are 1.B, 2.A, 2.C, 3.A and 3.B. Neither the AP Physics 1 nor the AP Physics 2 CED contains the phrase dot product, cross product, scalar product or vector product in any spelling, and neither of those booklets prints a VECTORS box; both Physics C booklets print one on their final appendix page, carrying A dot B = AB cosine theta and the magnitude of A cross B = AB sine theta. The AP Physics 2 sheet does write magnetic flux with a dot in its MAGNETISM box, giving both the vector form and the equivalent cosine form on consecutive lines.

One asks how much is along, the other how much is across

Two vectors, two ways to multiply them, and the two answers are not even the same kind of thing.

The dot product asks: how much of $\vec{A}$ lies along $\vec{B}$ ? The answer is a single number, $\vec{A} \cdot \vec{B} = AB\cos\theta$ , with no direction attached.
The cross product asks: how much of $\vec{A}$ lies perpendicular to $\vec{B}$ ? The answer is a vector of magnitude $AB\sin\theta$ pointing perpendicular to both.

AP Physics C: Mechanics states both, in two different units, and the wording of each is worth having exactly.

EK 3.2.A.3.i, in the work topic: the dot product between two vectors, $\vec{A}$ and $\vec{B}$ , results in a scalar quantity of magnitude $\vec{A} \cdot \vec{B} = AB\cos\theta$ .

EK 5.3.B.2.i, in the torque topic: the cross-product between two vectors, $\vec{A}$ and $\vec{B}$ , results in a vector quantity of magnitude $AB\sin\theta$ . EK 5.3.B.2.ii adds that the direction of the resulting vector is perpendicular to both $\vec{A}$ and $\vec{B}$ and therefore normal to the plane they define, and EK 5.3.B.2.iii says that direction can be qualitatively determined by applying the appropriate right-hand rule.

Scalar quantity against vector quantity. The CED puts those words in the two statements deliberately, and they are the whole difference. Everything else on this page is a consequence: which physical quantities use which product, why work has no direction and torque does, and which trigonometric function belongs in a given problem.

Where the two products are printed, and which courses ever name them, is covered in section seven. The AP Physics C vectors box reference carries the transcription of the printed table itself; this page is about which product a piece of physics needs and why.

Dot product vs cross product, side by side

Question you are asking	Dot product $\vec{A} \cdot \vec{B}$	Cross product $\vec{A} \times \vec{B}$
CED essential knowledge	3.2.A.3.i	5.3.B.2.i to 5.3.B.2.iii
What it returns	A scalar	A vector
Magnitude	$AB\cos\theta$	$AB\sin\theta$
Direction of the result	None	Perpendicular to both, by the right-hand rule
Largest when the vectors are	Parallel, $\theta = 0$	Perpendicular, $\theta = 90^\circ$
Zero when the vectors are	Perpendicular	Parallel or antiparallel
Can the result be negative	Yes, for $\theta > 90^\circ$	Not a magnitude; the direction reverses instead
Order of the two vectors	Does not matter	Reverses the result's direction
In components, two dimensions	$A_xB_x + A_yB_y$	$A_xB_y - A_yB_x$ , out of the plane
Geometric picture	Projection of one onto the other	Area of the parallelogram they span
Physics it builds	Work, flux, potential difference, power	Torque, angular momentum, magnetic force
Printed on the sheet	VECTORS box, both Physics C booklets	VECTORS box, both Physics C booklets

Four rows carry the page.

The largest-when and zero-when rows are exact opposites, and that is the fastest way to pick the right product. If lining the two vectors up makes the effect biggest, you want a cosine and a dot product. If lining them up kills the effect entirely, you want a sine and a cross product. Push a box along the direction it moves and you do the most work; pull a wrench straight along its handle and you get no torque at all.

The negative row is a real asymmetry. A dot product genuinely goes negative when the angle exceeds $90$ degrees, and that sign carries physics: negative work, or negative flux. A cross product's magnitude is never negative, because $\sin\theta \geq 0$ for $0 \leq \theta \leq 180^\circ$ ; what changes is the direction of the resulting vector.

The order row is the one that costs marks. $\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$ , always. But $\vec{B} \times \vec{A}$ points opposite to $\vec{A} \times \vec{B}$ , so torque written as $\vec{F} \times \vec{r}$ points the wrong way. The CED prints it as $\vec{\tau} = \vec{r} \times \vec{F}$ , position first.

The geometry row explains the two trigonometric functions in one picture. Draw the parallelogram spanned by $\vec{A}$ and $\vec{B}$ : its base is $A$ , its height is $B\sin\theta$ , and its area is $AB\sin\theta$ , the cross product magnitude. The dot product is the other projection, $B\cos\theta$ , times $A$ . Between them the two products account for both components of $\vec{B}$ relative to $\vec{A}$ , which is why $(\vec{A}\cdot\vec{B})^2 + \lvert \vec{A}\times\vec{B} \rvert^2 = (AB)^2$ . Worked example one checks that identity numerically.

Why work is a scalar and torque is a vector

This is the payoff, and it is not an analogy. The kind of quantity work is, and the kind torque is, are decided by which product each is built from.

Work is a dot product, so work is a scalar. AP Physics C: Mechanics EK 3.2.A.3 gives the general definition as $W = \displaystyle\int_a^b \vec{F}(r) \cdot d\vec{r}$ , with the integral taken over the path from point $a$ to point $b$ , and it is that centered dot that decides everything downstream. EK 3.2.A.3.ii says why the projection is the right operation: only the component of the force exerted on a system that is parallel to the displacement of the point of application of the force will change the system's total energy. A force at right angles to the motion changes nothing about the energy, which EK 3.2.A.3.iv states from the other side: the perpendicular component can change the direction of the system's motion without changing the system's kinetic energy. So work asks how much of the force is along the motion, gets a cosine, and comes out as a number of joules with no direction. There is no such thing as work pointing north.

Torque is a cross product, so torque is a vector. EK 5.3.B.2 gives $\vec{\tau} = \vec{r} \times \vec{F}$ , and EK 5.3.A.1 says why the perpendicular projection is the right one: torque results only from the force component perpendicular to the position vector from the axis of rotation to the point of application of the force. A force pointing straight at the axis produces no turning whatsoever, which is the exact mirror of the work statement. So torque asks how much of the force is across the position vector, gets a sine, and comes out as a vector along the axis of rotation, because that is the only direction that describes a rotation in the plane.

Hold those two next to each other:

	Work	Torque
Built from	$\vec{F} \cdot d\vec{r}$	$\vec{r} \times \vec{F}$
The useful force component is	Parallel to the displacement	Perpendicular to the position vector
Trigonometric factor	$\cos\theta$	$\sin\theta$
Zero when the force is	Perpendicular to the motion	Along the line to the axis
Result	A scalar, in joules	A vector, in newton metres
CED statement of the projection	EK 3.2.A.3.ii	EK 5.3.A.1

Same force, same angle, and the two answers are complements of each other. Worked example two takes one $25$ N force at $30$ degrees and produces a work and a torque from it, and as the angle sweeps from $0$ to $90$ degrees one grows exactly as the other shrinks.

A sentence worth writing down: the units do not tell you which is which. A joule and a newton metre are the same combination of base units. What distinguishes them is that one came out of a projection and the other out of a perpendicular, and that is a fact about the operation, not about the dimensions.

The catalogue: which AP equations use which product

Once you can read the two symbols, whole regions of the equation sheet become sortable at a glance. Every line below is printed in the AP Physics C booklets.

Dot products, all asking how much of one thing lies along another:

Printed line	What it is projecting onto what
$W = \displaystyle\int_a^b \vec{F} \cdot d\vec{r}$	Force onto displacement
$\Delta U = -\displaystyle\int_a^b \vec{F}_{\text{cf}}(r) \cdot d\vec{r}$	Conservative force onto displacement
$\Phi_E = \displaystyle\int \vec{E} \cdot d\vec{A}$	Electric field onto the area vector
$\displaystyle\oint \vec{E} \cdot d\vec{A} = \dfrac{q_{\text{enc}}}{\varepsilon_0}$	The same, over a closed surface
$\Delta V = -\displaystyle\int_a^b \vec{E} \cdot d\vec{r}$	Electric field onto displacement
$\Phi_B = \displaystyle\int \vec{B} \cdot d\vec{A}$	Magnetic field onto the area vector
$\displaystyle\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\text{enc}}$	Magnetic field onto a path element

Cross products, all asking how much of one thing lies perpendicular to another:

Printed line	What is perpendicular to what
$\vec{\tau} = \vec{r} \times \vec{F}$	Force perpendicular to the position vector
$\vec{L} = \vec{r} \times \vec{p} = I\vec{\omega}$	Momentum perpendicular to the position vector
$\vec{F}_B = q\left(\vec{v} \times \vec{B}\right)$	Velocity perpendicular to the magnetic field
$\vec{F}_B = \displaystyle\int I\, d\vec{\ell} \times \vec{B}$	Current element perpendicular to the field
$d\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{I\, d\vec{\ell} \times \hat{r}}{r^2}$	Current element perpendicular to the direction to the point

The pattern is not decoration. Flux is a dot product because flux counts field lines that go through a surface, and a field line running along the surface goes through nothing. EK 8.5.A.2 in AP Physics C: Electricity and Magnetism writes electric flux for a constant field as $\Phi_E = \vec{E} \cdot \vec{A}$ , EK 8.5.A.2.i defines the area vector's direction as perpendicular to the plane of the surface and outward from a closed surface, and EK 8.5.A.2.ii says the sign of flux is given by the dot product of the electric field vector and the area vector. EK 13.1.A.1.ii repeats that last statement for magnetic flux. So the negative-dot-product case is not a curiosity here; it is how inward flux gets its minus sign.

The magnetic force is a cross product because a charge moving along the field feels nothing. EK 12.2.B.1 states it in those terms: a magnetic field will exert a force on a charged object moving within that field, with magnitude and direction that depend on the cross-product of the charge's velocity and the magnetic field, and the relevant equation is $\vec{F}_B = q(\vec{v} \times \vec{B})$ .

That one has a consequence worth deriving, because it links the two products in a single sentence. Since $\vec{v} \times \vec{B}$ is perpendicular to $\vec{v}$ by EK 5.3.B.2.ii, the magnetic force is always perpendicular to the velocity, so the dot product $\vec{F}_B \cdot \vec{v}$ is zero, so a magnetic field can never do work on a moving charge. A cross product feeding a dot product that has to vanish. Worked example three runs it.

The case that separates them: one force, two questions, one angle

A $25$ N force acts at $30$ degrees to a $2.0$ m position vector from a pivot, and the object's point of application moves $2.0$ m along that same position vector. Ask the two questions.

Angle between the vectors	Work, $Fd\cos\theta$	Torque magnitude, $rF\sin\theta$
$0^\circ$	$50.0$ J	$0$
$30^\circ$	$43.3$ J	$25.0\ \text{N}\cdot\text{m}$
$60^\circ$	$25.0$ J	$43.3\ \text{N}\cdot\text{m}$
$90^\circ$	$0$	$50.0\ \text{N}\cdot\text{m}$
$180^\circ$	$-50.0$ J	$0$

Read the two columns against each other. Every configuration that is best for one is worst for the other. The $0$ degree row does maximum work and produces no turning; the $90$ degree row produces maximum torque and does no work along the radius. The two columns are the two components of the same force, sorted by the two products.

The $180$ degree row shows the other asymmetry. The work has gone negative, which is a physical statement: energy is being taken out of the system. The torque has gone back to zero rather than negative, because $\sin 180^\circ = 0$ ; a force pulling directly away from the pivot turns nothing, just as one pushing directly toward it turns nothing. The dot product changes sign at $90$ degrees; the cross product changes direction at $180$ degrees.

One more separation, and it is the one that decides what you can do with the answer. Two works add as numbers: total work is $43.3 + 25.0 = 68.3$ J with no diagram required. Two torques do not add as numbers unless they share an axis and a sense, because each is a vector. On a flat problem with everything in one plane the two directions collapse to clockwise and counterclockwise and the vector nature hides, which is exactly why AP Physics 1 can handle torque with signed magnitudes and never mention a cross product. In three dimensions that shortcut fails, and AP Physics C introduces the cross-product to replace it.

Where the confusion costs a mark

Each of these is a specific scoring error.

Using $\sin$ where the physics wants $\cos$ , or the reverse. The one-line test: does lining the two vectors up make the quantity bigger or kill it? Bigger means cosine and a dot product. Killed means sine and a cross product.
Measuring flux angles from the surface instead of from its normal. EK 8.5.A.2.i defines the area vector as perpendicular to the plane of the surface, so the angle in $\Phi_E = EA\cos\theta$ is measured from that normal. A field at $30$ degrees to the surface is at $60$ degrees to the normal, and using the wrong one changes the answer by $42$ percent, which worked example three computes.
Writing torque as $\vec{F} \times \vec{r}$ . The cross product reverses when you swap the order, so this points along the axis in the wrong direction. The printed form is $\vec{\tau} = \vec{r} \times \vec{F}$ .
Giving work a direction. EK 3.2.A.3.i calls the dot product a scalar quantity. Work in joules with a direction attached is a category error, and it is worth catching before it produces an energy equation with vectors in it.
Reporting a torque or a magnetic force with only a magnitude. The other half of the same mistake. EK 5.3.B.2.i calls the cross-product a vector quantity, and EK 5.3.B.2.iii says the direction can be determined with the appropriate right-hand rule, so an answer without a direction is incomplete.
Assuming the magnetic force does work. It cannot, because $q(\vec{v} \times \vec{B})$ is perpendicular to $\vec{v}$ . A magnetic field changes a charge's direction and never its speed.
Taking the dot product of vectors that are not both vectors. The area vector, the path element $d\vec{\ell}$ and the position vector all have to be treated as vectors with defined directions before either product means anything.
Assuming a zero dot product means the vectors are zero. It means they are perpendicular. Likewise a zero cross product means they are parallel or antiparallel, not absent.
Looking for the vector products on an algebra-based sheet. Neither AP Physics 1 nor AP Physics 2 prints a vectors box, and neither of those CEDs uses the phrases dot product or cross product anywhere. If a question in those courses needs a projection, it prints the cosine or the sine explicitly, as in $W = F_{\parallel}d = Fd\cos\theta$ and $\tau = r_{\perp}F = rF\sin\theta$ .

Which courses use them, and where the products are printed

This is where the algebra-based and calculus-based tracks part company, and the split is straightforward.

AP Physics C uses both products formally. The phrase dot product appears in the AP Physics C: Mechanics Course and Exam Description at EK 3.2.A.3.i, and cross-product at EK 5.3.B.2.i, 5.3.B.2.ii and 5.3.B.2.iii. In the AP Physics C: Electricity and Magnetism CED, dot product appears at EK 8.5.A.2.ii for electric flux, at EK 13.1.A.1.ii for magnetic flux and in the electric potential topic, and cross-product at EK 12.2.B.1 for the magnetic force.

Neither algebra-based course uses either phrase. Searching the full text of the AP Physics 1 and AP Physics 2 Course and Exam Descriptions returns no occurrence of dot product, cross product, scalar product or vector product, in any spelling or hyphenation. That is a scope decision, not an omission: those courses do the same projections with explicit trigonometry.

On the sheets, the two products are printed in a VECTORS box that only the Physics C booklets carry. It sits on the final appendix page beside the CALCULUS and IDENTITIES boxes, and it prints

\vec{A} \cdot \vec{B} = AB\cos\theta \qquad\qquad \lvert \vec{A} \times \vec{B} \rvert = AB\sin\theta

along with a unit-vector expansion and two statements of vector addition. The AP Physics 1 and AP Physics 2 appendices have no vectors box at all; they carry the geometry box and the trigonometric values table, and no table of vector operations. The AP Physics C vectors box reference transcribes all five printed lines.

Two details are worth having exactly, because both are easy to get slightly wrong.

First, the printed cross-product line carries magnitude bars and the essential knowledge statement does not. The sheet prints $\lvert \vec{A} \times \vec{B} \rvert = AB\sin\theta$ ; EK 5.3.B.2.i writes the same relation and puts the word magnitude into the sentence instead. They say the same thing, and the sheet's version is the safer one to quote.

Second, the AP Physics 2 sheet does print a dot, once, without ever defining it. In its MAGNETISM box, magnetic flux appears as $\Phi_B = \vec{B} \cdot \vec{A}$ on one line, with $\Phi_B = \lvert \vec{B} \rvert \cos\theta \lvert \vec{A} \rvert$ on the next. The second line is doing the work of the missing vectors box: it spells out what the dot means for a student whose course never named the operation.

For the CED framing, see Topic 3.2, Work and Topic 5.3, Torque in C: Mechanics, and Topic 8.5, Electric Flux and Topic 12.2, Magnetism and Moving Charges in C: Electricity and Magnetism. The algebra-based versions of the same physics are at Topic 3.2, Work and Topic 5.3, Torque. How to calculate torque has the routine, and the full C: Mechanics sheet has every line.

The same two vectors, put through both products

Vector $\vec{A}$ has magnitude $6.0$ and points along the $x$ axis. Vector $\vec{B}$ has magnitude $4.0$ and lies in the $xy$ plane at $60$ degrees above the $x$ axis. Find $\vec{A} \cdot \vec{B}$ and the magnitude and direction of $\vec{A} \times \vec{B}$ . Check both against a component calculation, and check that the two results satisfy the identity relating them to $AB$ .

Dot product from EK 3.2.A.3.i. $\vec{A} \cdot \vec{B} = AB\cos\theta = (6.0)(4.0)\cos 60^\circ = (24)(0.500) = 12.0$ . A pure number, no units of direction, no arrow. The trigonometric values table printed with every AP booklet gives $\cos 60^\circ = 1/2$ exactly.
Cross product magnitude from EK 5.3.B.2.i. $\lvert \vec{A} \times \vec{B} \rvert = AB\sin\theta = (24)\sin 60^\circ = (24)(0.8660) = 20.8$ .
Cross product direction from EK 5.3.B.2.ii and 5.3.B.2.iii. The result is perpendicular to both vectors and therefore normal to the $xy$ plane. Point the fingers of the right hand along $\vec{A}$ , curl them toward $\vec{B}$ , and the thumb points out of the page, that is along $+z$ .
Components of $\vec{B}$ . $B_x = 4.0\cos 60^\circ = 2.00$ and $B_y = 4.0\sin 60^\circ = 3.464$ . And $\vec{A} = (6.0, 0)$ .
Check the dot product in components. $A_xB_x + A_yB_y = (6.0)(2.00) + (0)(3.464) = 12.0$ . Matches.
Check the cross product in components. In two dimensions the out-of-plane component is $A_xB_y - A_yB_x = (6.0)(3.464) - (0)(2.00) = 20.8$ , positive, so out of the page. Matches both the magnitude and the right-hand-rule direction.
Check the identity. $(\vec{A} \cdot \vec{B})^2 + \lvert \vec{A} \times \vec{B} \rvert^2 = (24\cos 60^\circ)^2 + (24\sin 60^\circ)^2 = 576(0.250) + 576(0.750) = 144 + 432 = 576$ , and $(AB)^2 = (24)^2 = 576$ . They agree, because $\cos^2\theta + \sin^2\theta = 1$ , which is printed in the IDENTITIES box of the same Physics C appendix page as the vectors box.
Read what the identity says. The two products split $\vec{B}$ into its along part and its across part relative to $\vec{A}$ , and nothing is left over. Any question about two vectors is answered by one of them or by both.

$\vec{A} \cdot \vec{B} = 12.0$ , a scalar. $\lvert \vec{A} \times \vec{B} \rvert = 20.8$ , a vector pointing out of the $xy$ plane along $+z$ . Component calculations reproduce both, and $(12.0)^2 + (20.8)^2 = 576 = (24)^2$ , confirming that the two products account for the whole of one vector relative to the other.

One force at 30 degrees, giving a scalar work and a vector torque

A $25$ N force is applied at the end of a $2.0$ m rod, at $30$ degrees to the rod. The rod is pivoted at the other end. Find the torque about the pivot. Separately, suppose the point of application moves $2.0$ m along the rod's own direction while the same force acts at the same $30$ degrees to that displacement; find the work done. Then repeat both at $0$ , $60$ , $90$ and $180$ degrees, and say what each answer means physically.

Torque at $30$ degrees. The printed relation is $\tau = r_{\perp}F = rF\sin\theta$ , the magnitude of $\vec{\tau} = \vec{r} \times \vec{F}$ . So $\tau = (2.0)(25)\sin 30^\circ = (50)(0.500) = 25.0\ \text{N}\cdot\text{m}$ , directed along the axis, out of the plane of the rod and force by the right-hand rule.
Work at $30$ degrees. The printed relation is $W = F_{\parallel}d = Fd\cos\theta$ , the constant-force form of $W = \int \vec{F} \cdot d\vec{r}$ . So $W = (25)(2.0)\cos 30^\circ = (50)(0.8660) = 43.3\ \text{J}$ , a scalar.
At $0$ degrees. $W = (50)(1) = 50.0\ \text{J}$ , the maximum, because the force is entirely along the motion. $\tau = (50)(0) = 0$ , because a force pointing straight along the rod passes through the pivot and turns nothing. This is EK 5.3.A.1: torque results only from the force component perpendicular to the position vector.
At $60$ degrees. $W = (50)(0.500) = 25.0\ \text{J}$ and $\tau = (50)(0.8660) = 43.3\ \text{N}\cdot\text{m}$ . The two numbers from the $30$ degree case have swapped, because $\cos 60^\circ = \sin 30^\circ$ and $\sin 60^\circ = \cos 30^\circ$ .
At $90$ degrees. $W = 0$ and $\tau = 50.0\ \text{N}\cdot\text{m}$ , the maximum. EK 3.2.A.3.iv describes this case exactly: the perpendicular component can change the direction of the system's motion without changing its kinetic energy.
At $180$ degrees. $W = (50)(-1) = -50.0\ \text{J}$ , negative work, so energy is being removed. $\tau = (50)(0) = 0$ again: pulling straight away from the pivot turns nothing, just as pushing straight toward it does not.
Note what the two zero-torque rows have in common and the two extreme-work rows do not. Torque hits zero at both $0$ and $180$ degrees and never goes negative in magnitude; it reverses direction instead. Work runs continuously from $+50.0$ J down through zero to $-50.0$ J. The dot product carries a sign; the cross product carries a direction.
Units, as a final check on which is which. Both answers are $\text{N}\cdot\text{m}$ in base units, and a joule is a newton metre. The dimensions cannot tell them apart, and the reason they are different quantities is that one came from a projection and one from a perpendicular.

At $30$ degrees the torque is $25.0\ \text{N}\cdot\text{m}$ as a vector along the axis, and the work is $43.3$ J as a scalar. Sweeping the angle gives work of $50.0$ , $43.3$ , $25.0$ , $0$ and $-50.0$ J at $0$ , $30$ , $60$ , $90$ and $180$ degrees, and torque magnitudes of $0$ , $25.0$ , $43.3$ , $50.0$ and $0\ \text{N}\cdot\text{m}$ at the same angles. Every configuration best for one is worst for the other.

Electric flux and magnetic force, and the work a magnetic field cannot do

A uniform electric field of magnitude $350$ N/C passes through a flat surface of area $0.020\ \text{m}^2$ whose normal is at $30$ degrees to the field. Find the electric flux. Then a proton with charge $1.60 \times 10^{-19}$ C moves at $4.0 \times 10^5$ m/s at $30$ degrees to a magnetic field of $0.25$ T; find the magnitude of the magnetic force on it and say what work that force does over any interval.

Electric flux, using the dot product. EK 8.5.A.2 gives $\Phi_E = \vec{E} \cdot \vec{A}$ for a field constant across the area, so $\Phi_E = EA\cos\theta$ with $\theta$ measured from the area vector, which EK 8.5.A.2.i defines as perpendicular to the plane of the surface.
$\Phi_E = (350)(0.020)\cos 30^\circ = (7.00)(0.8660) = 6.06\ \text{N}\cdot\text{m}^2/\text{C}$ . A scalar, and positive because the angle is under $90$ degrees, which is EK 8.5.A.2.ii's statement that the sign of flux is given by the dot product of the field vector and the area vector.
The trap, priced. Suppose the $30$ degrees had been read as the angle to the surface rather than to its normal, and a sine used: $(7.00)\sin 30^\circ = 3.50\ \text{N}\cdot\text{m}^2/\text{C}$ . That is $42$ percent below the correct value. The two functions are not close at $30$ degrees, so this error never hides in the rounding.
Magnetic force, using the cross product. EK 12.2.B.1 gives $\vec{F}_B = q(\vec{v} \times \vec{B})$ , whose magnitude by EK 5.3.B.2.i is $qvB\sin\theta$ with $\theta$ the angle between $\vec{v}$ and $\vec{B}$ .
$\lvert \vec{F}_B \rvert = (1.60 \times 10^{-19})(4.0 \times 10^5)(0.25)\sin 30^\circ$ . Build it up: $(1.60 \times 10^{-19})(4.0 \times 10^5) = 6.4 \times 10^{-14}$ ; times $0.25$ gives $1.6 \times 10^{-14}$ ; times $0.500$ gives $8.0 \times 10^{-15}\ \text{N}$ .
Its direction. By EK 5.3.B.2.ii the result of a cross-product is perpendicular to both input vectors, so this force is perpendicular to the proton's velocity and to the magnetic field, and EK 5.3.B.2.iii says the right-hand rule fixes which of the two perpendicular directions it is.
Now the work. Work is the dot product of force with displacement, and the displacement is along the velocity. Since $\vec{F}_B$ is perpendicular to $\vec{v}$ at every instant, $\vec{F}_B \cdot \vec{v} = F_B v \cos 90^\circ = 0$ , so the power delivered is zero and the work over any interval is zero.
Read that result. A cross product produced a vector that a dot product then had to annihilate. The magnetic field changes the proton's direction continuously and its speed not at all, which is why charged particles in uniform magnetic fields travel in circles at constant speed. EK 3.2.A.3.iv described this case in the work topic long before magnetism appeared: the perpendicular component of a force can change the direction of a system's motion without changing its kinetic energy.

The electric flux is $6.06\ \text{N}\cdot\text{m}^2/\text{C}$ , a scalar, positive. Reading the angle from the surface instead of from the normal would have given $3.50$ , an error of $42$ percent. The magnetic force on the proton is $8.0 \times 10^{-15}$ N, a vector perpendicular to both the velocity and the field, and it does exactly zero work no matter how long the proton travels, because a cross-product result is perpendicular to its inputs.

Frequently asked questions

What is the difference between the dot product and the cross product?

The dot product multiplies two vectors and returns a scalar, a single number equal to A times B times the cosine of the angle between them, with no direction attached. The cross product returns a vector whose magnitude is A times B times the sine of that angle and whose direction is perpendicular to both input vectors, fixed by the right-hand rule. The AP Physics C: Mechanics CED states both, at essential knowledge 3.2.A.3.i for the dot product and essential knowledge 5.3.B.2.i for the cross product, and it is careful to say scalar quantity in the first and vector quantity in the second. The dot product asks how much of one vector lies along the other; the cross product asks how much lies perpendicular to it.

Why is work a scalar and torque a vector?

Because work is built from a dot product and torque from a cross product. Work is the integral of force dotted with displacement, and a dot product returns a scalar, so work has a magnitude and a sign but no direction. Torque is the position vector crossed with the force, and a cross product returns a vector, so torque points along the axis of rotation. The underlying physics matches: essential knowledge 3.2.A.3.ii says only the force component parallel to the displacement changes a system's total energy, while essential knowledge 5.3.A.1 says torque results only from the force component perpendicular to the position vector. Parallel and perpendicular, cosine and sine, scalar and vector.

How do you know whether to use sine or cosine?

Ask whether lining the two vectors up makes the quantity bigger or destroys it. If pointing them the same way maximises the effect, the quantity is a projection and you want cosine and a dot product: work, flux, potential difference. If pointing them the same way kills the effect, the quantity depends on the perpendicular part and you want sine and a cross product: torque, angular momentum, magnetic force. A force pushing a box along its motion does the most work and a force pulling straight along a wrench handle produces no torque, which is the two rules in one sentence each.

Is the cross product on the AP Physics equation sheet?

Yes, on both AP Physics C booklets. The magnitude of A cross B equals A B sine theta is printed in a box headed VECTORS on the final appendix page, directly under the dot product line, in both the C: Mechanics and the C: Electricity and Magnetism Course and Exam Descriptions. It is often reported as missing because most machine-readable transcriptions of the AP sheets cover only the physics equations and constants boxes and leave out the geometry, trigonometry, vector, calculus and identity tables. Neither AP Physics 1 nor AP Physics 2 has a vectors box on its appendix pages at all.

Do AP Physics 1 and AP Physics 2 use dot and cross products?

Not by name and not as operations. Searching the full text of both Course and Exam Descriptions turns up no occurrence of dot product, cross product, scalar product or vector product in any spelling. The algebra-based courses do the same projections with explicit trigonometry instead, printing work as F d cosine theta and torque as r F sine theta on their equation sheets, and handling torque direction as clockwise or counterclockwise rather than as a vector along an axis. That shortcut works because those courses keep rotations in a single plane. AP Physics C introduces the products where three dimensions make the shortcut fail.

Why can a magnetic field never do work on a moving charge?

Because the magnetic force is a cross product of the velocity with the field, and the result of a cross product is perpendicular to both of its inputs. Essential knowledge 12.2.B.1 in AP Physics C: Electricity and Magnetism gives the force as q times v cross B, and essential knowledge 5.3.B.2.ii in C: Mechanics states the perpendicularity. So the magnetic force is always at right angles to the velocity, and work is the dot product of force with displacement, which is zero at right angles. The field changes the charge's direction continuously and its speed not at all, which is why a charged particle in a uniform magnetic field moves in a circle at constant speed.

Why is electric flux a dot product?

Because flux counts how much field passes through a surface, and field running along the surface passes through nothing. Essential knowledge 8.5.A.2 in AP Physics C: Electricity and Magnetism writes the flux of a constant field as E dotted with the area vector, and essential knowledge 8.5.A.2.i defines that area vector as perpendicular to the plane of the surface and pointing outward from a closed surface. The angle in the cosine is therefore measured from the normal, not from the surface, which is a common error on flux questions. Essential knowledge 8.5.A.2.ii adds that the sign of the flux comes from the dot product, which is how flux entering a closed surface gets its minus sign.