Vector (Cross) Product

Stand at a door. Push it at the handle, perpendicular to the surface. It swings open easily. Now push at the same point but along the door's edge, parallel to the surface. Nothing happens — the door does not budge. Same force, same point, but a completely different result. The difference is the angle between the force and the lever arm. Physics needs an operation that captures this angle-dependent, direction-producing combination of two vectors. That operation is the cross product.

The dot product takes two vectors and returns a scalar — a single number. The cross product does something bolder: it takes two vectors and returns a new vector, one that points in a direction neither of the original vectors was pointing. The magnitude of this new vector depends on the sine of the angle between them, not the cosine. And the direction? It is perpendicular to both — sticking straight out of the plane the two vectors define.

This is the operation that gives physics torque, angular momentum, and the magnetic force on a moving charge. Without the cross product, none of these quantities could be defined.

The geometric definition

Take two vectors \vec{A} and \vec{B} with an angle \theta between them (measured as the smaller angle, so 0 \leq \theta \leq 180°). The cross product \vec{A} \times \vec{B} is defined by two rules — one for the magnitude, one for the direction.

Magnitude:

Why sine and not cosine? The dot product uses \cos\theta because it measures how much two vectors point along each other — the "overlap." The cross product uses \sin\theta because it measures how much they point across each other — the "perpendicular component." When two vectors are parallel (\theta = 0°), there is no "across" — and \sin 0° = 0, so the cross product vanishes. When they are perpendicular (\theta = 90°), the "across" is maximum — and \sin 90° = 1.

Direction: perpendicular to both \vec{A} and \vec{B}, determined by the right-hand rule.

The right-hand rule

The plane defined by \vec{A} and \vec{B} has two perpendicular directions — "up" and "down," so to speak. Which one does \vec{A} \times \vec{B} choose? The right-hand rule settles this.

Here is how to use it: hold your right hand flat. Point your fingers along \vec{A}. Now curl your fingers toward \vec{B} (through the smaller angle between them). Your thumb — sticking out of the plane — points in the direction of \vec{A} \times \vec{B}.

Try it right now with something concrete. Lay your phone flat on the table — that is the plane of \vec{A} and \vec{B}. If \vec{A} points toward the right edge and \vec{B} points toward the top edge, curl your right hand's fingers from right to top: your thumb points upward, out of the screen. That is the direction of \vec{A} \times \vec{B}.

A quick way to remember: if the rotation from \vec{A} to \vec{B} looks counterclockwise when viewed from the tip of the cross product, you have the correct direction.

Reading the definition. The factor |\vec{A}||\vec{B}| gives you the product of the two magnitudes — how "big" the two vectors are. The factor \sin\theta controls how much of that product survives: full product at \theta = 90° (perpendicular), zero at \theta = 0° or 180° (parallel or anti-parallel). And \hat{n} is the direction — perpendicular to both, chosen by the right-hand rule.

Compare this with the dot product: \vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta. The dot product tells you "how aligned are these vectors?" The cross product tells you "how perpendicular are these vectors?" They are complementary — together they capture all the geometric information about the angle between two vectors.

Building the cross product from unit vectors

To use the cross product in calculations, you need the component form. Start from the three unit vectors \hat{i}, \hat{j}, \hat{k} along the x, y, z axes. These are mutually perpendicular, so the angle between any two different unit vectors is 90°.

Apply the definition to pairs of unit vectors.

Same-direction pairs: \hat{i} \times \hat{i}, \hat{j} \times \hat{j}, \hat{k} \times \hat{k}.

Why: the angle between a vector and itself is 0°. Since \sin 0° = 0, the cross product vanishes. Any vector crossed with itself gives the zero vector. This makes physical sense — a single vector defines no plane, so there is no perpendicular direction to pick.

Different-direction pairs: \hat{i} \times \hat{j}, \hat{j} \times \hat{k}, \hat{k} \times \hat{i}.

The right-hand rule: point fingers along \hat{i} (the x-axis), curl toward \hat{j} (the y-axis). Your thumb points along the z-axis — that is \hat{k}.

Why: the x and y axes define the xy-plane, and the perpendicular to that plane is the z-axis. The right-hand rule picks the positive z-direction.

By the same logic, cycling through i \to j \to k \to i:

Why: this cyclic pattern (i \to j \to k \to i) follows from the right-hand rule applied to each pair. Going "forward" in the cycle gives a positive result; going "backward" gives a negative result.

And in the reverse order:

Why: swapping the order reverses the curl direction of the right hand, so the thumb flips — the result picks up a minus sign. This is the anti-commutative property at work.

A mnemonic: write i, j, k in a cycle. Going clockwise (i \to j \to k) gives positive. Going counter-clockwise (k \to j \to i) gives negative.

The component form — the determinant formula

Now you are ready to derive the general formula. Write \vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k} and \vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}.

Step 1. Expand the cross product using the distributive property (which the cross product obeys — proved in the properties section below).

Why: the cross product distributes over addition, just like ordinary multiplication. So you can expand this product term by term — nine terms in total.

Step 2. Multiply out all nine terms.

Why: each component of \vec{A} crosses with each component of \vec{B}. The scalar coefficients (A_xB_y etc.) pull out in front because the cross product is linear in each argument.

Step 3. Substitute the unit vector cross products from the previous section. The three "same" terms vanish (\hat{i} \times \hat{i} = \vec{0}, etc.), leaving six surviving terms.

Step 4. Group by unit vector.

Why: collect the \hat{i} terms together, the \hat{j} terms together, and the \hat{k} terms together. Each component is a "mini cross product" of the other two component pairs — a pattern that becomes clearer in the determinant form.

This result can be written compactly as a determinant:

Why: expanding this 3 \times 3 determinant along the first row produces exactly the three components derived above. The \hat{i} component is the 2 \times 2 determinant A_yB_z - A_zB_y, the \hat{j} component picks up a minus sign giving A_zB_x - A_xB_z, and the \hat{k} component is A_xB_y - A_yB_x.

This determinant is a mnemonic, not a "real" determinant (the first row contains vectors, not numbers), but it always gives the correct answer and is the fastest way to compute a cross product in practice.

Properties of the cross product

Anti-commutativity: \vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})

Swap \vec{A} and \vec{B} in the geometric definition. The magnitude stays the same: |\vec{B}||\vec{A}|\sin\theta = |\vec{A}||\vec{B}|\sin\theta. But the right-hand rule reverses: curl from \vec{B} to \vec{A} instead of \vec{A} to \vec{B}, and your thumb flips. The result is the same vector with the opposite sign.

Why this matters: the dot product is commutative (\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}), but the cross product is not. Order matters. If you swap the two vectors, you get the negative of what you had before. This has real physical consequences — swapping the order of \vec{r} and \vec{F} in the torque formula \vec{\tau} = \vec{r} \times \vec{F} would reverse the torque direction.

You can verify this from the component form directly. Swap A and B in the determinant:

Each component is the negative of the corresponding component of \vec{A} \times \vec{B}. So \vec{B} \times \vec{A} = -(\vec{A} \times \vec{B}).

Distributive property: \vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C}

The cross product distributes over vector addition. This is what allowed the component-by-component expansion in the derivation above. The proof is most easily done using the component form — each component of \vec{A} \times (\vec{B} + \vec{C}) expands to the sum of the corresponding components of \vec{A} \times \vec{B} and \vec{A} \times \vec{C}.

Zero when parallel: \vec{A} \times \vec{B} = \vec{0} when \vec{A} \parallel \vec{B}

If two vectors are parallel (\theta = 0°) or anti-parallel (\theta = 180°), then \sin\theta = 0, so |\vec{A} \times \vec{B}| = 0. The cross product is the zero vector.

Why: parallel vectors define no unique plane — there are infinitely many planes containing a single line. Since there is no unique plane, there is no unique perpendicular direction, and the cross product cannot pick one. The mathematics resolves this by making the magnitude zero.

Corollary: \vec{A} \times \vec{A} = \vec{0} for any vector \vec{A}. Every vector is parallel to itself.

Scalar multiplication: (\lambda\vec{A}) \times \vec{B} = \lambda(\vec{A} \times \vec{B})

A scalar factor pulls out of the cross product. This follows directly from the magnitude formula: |\lambda\vec{A}||\vec{B}|\sin\theta = |\lambda|\,|\vec{A}||\vec{B}|\sin\theta, with the sign of \lambda determining whether the direction flips.

NOT associative: \vec{A} \times (\vec{B} \times \vec{C}) \neq (\vec{A} \times \vec{B}) \times \vec{C} in general

Unlike ordinary multiplication, the cross product is not associative. A quick counterexample: \hat{i} \times (\hat{i} \times \hat{j}) = \hat{i} \times \hat{k} = -\hat{j}, but (\hat{i} \times \hat{i}) \times \hat{j} = \vec{0} \times \hat{j} = \vec{0}. The two results are different, so associativity fails. When you see a triple cross product, the brackets matter.

The parallelogram area interpretation

The cross product has a beautiful geometric meaning beyond the definition: the magnitude |\vec{A} \times \vec{B}| equals the area of the parallelogram formed by \vec{A} and \vec{B}.

Here is why. Place \vec{A} and \vec{B} tail-to-tail. They form two sides of a parallelogram. The area of a parallelogram is base times height. Take |\vec{A}| as the base. The height is the perpendicular distance from the tip of \vec{B} to the line of \vec{A}, which is |\vec{B}|\sin\theta.

Why: the \sin\theta factor gives exactly the perpendicular component of \vec{B} relative to \vec{A}, which is the height of the parallelogram. This is not a coincidence — it is what the cross product was designed to capture.

Corollary — triangle area: Since a triangle is half a parallelogram, the area of the triangle formed by \vec{A} and \vec{B} is \frac{1}{2}|\vec{A} \times \vec{B}|. This is often the quickest way to find the area of a triangle when you know the position vectors of its vertices.

Applications — why physics needs the cross product

The cross product is not abstract mathematics — it is the language of rotation and perpendicular forces. Here are the three most important physical quantities that are defined as cross products.

Torque: \vec{\tau} = \vec{r} \times \vec{F}

Torque is the rotational effect of a force. When you push a door at the handle, you apply a force \vec{F} at a position \vec{r} from the hinge (the pivot). The torque about the pivot is:

Its magnitude is \tau = rF\sin\theta, where \theta is the angle between the position vector and the force. Maximum torque at \theta = 90° (push perpendicular to the door), zero torque at \theta = 0° (push along the door's edge).

This is why you push a door at the handle, not near the hinge: the handle is at the largest r, giving the greatest torque for the same force. And this is why you push perpendicular to the door (\theta = 90°, \sin\theta = 1) rather than at a shallow angle.

Angular momentum: \vec{L} = \vec{r} \times \vec{p}

The angular momentum of a particle about a point is \vec{L} = \vec{r} \times \vec{p}, where \vec{r} is the position vector from the point and \vec{p} = m\vec{v} is the linear momentum. A cricket ball spinning through the air has angular momentum about its centre. The Earth orbiting the Sun has angular momentum about the Sun. In each case, the cross product ensures \vec{L} points along the axis of rotation.

Magnetic force: \vec{F} = q\vec{v} \times \vec{B}

When a charged particle moves through a magnetic field, the force on it is:

This force is always perpendicular to both the velocity \vec{v} and the magnetic field \vec{B} — exactly what the cross product produces. If the charge moves parallel to the field, \sin\theta = 0 and there is no force. If it moves perpendicular to the field, the force is maximum. The cross product is the natural language for this physics.

Worked examples

Common confusions

• "The cross product is a scalar." No — the cross product is a vector. You may be thinking of the dot product, which produces a scalar. The cross product produces a vector with magnitude and direction. Its magnitude happens to equal the parallelogram area, but the cross product itself has a direction too — perpendicular to both inputs.

• "The direction of \vec{A} \times \vec{B} depends on the magnitudes." No — the direction depends only on the directions of \vec{A} and \vec{B} (via the right-hand rule), not their magnitudes. Doubling |\vec{A}| doubles the magnitude of the cross product but does not change its direction.

• "\sin\theta means the cross product is zero when vectors are perpendicular." This is backwards. \sin 90° = 1, so the cross product is maximum when the vectors are perpendicular. It is zero when they are parallel (\sin 0° = 0). You may be confusing this with the dot product, which uses \cos\theta and is zero when perpendicular.

• "The cross product is commutative, like the dot product." No — it is anti-commutative: \vec{A} \times \vec{B} = -(\vec{B} \times \vec{A}). Swapping the order reverses the sign. This is not just a mathematical curiosity — in physics, writing \vec{F} \times \vec{r} instead of \vec{r} \times \vec{F} for torque gives you the wrong direction.

• "I can take the cross product of two 2D vectors." Strictly, the cross product is defined for three-dimensional vectors. If your vectors lie in the xy-plane (so A_z = B_z = 0), the cross product has only a \hat{k} component: \vec{A} \times \vec{B} = (A_xB_y - A_yB_x)\hat{k}. This is perpendicular to the plane — which is why 2D torque and 2D angular momentum point "out of the page."

Where this leads next

• Torque — the full treatment of rotational force, including the torque due to multiple forces and the equilibrium condition \sum\vec{\tau} = \vec{0}.
• Angular Momentum — the rotational analogue of linear momentum, defined as \vec{L} = \vec{r} \times \vec{p}, and the conservation law that governs spinning objects.
• Magnetic Force on a Moving Charge — the Lorentz force \vec{F} = q\vec{v} \times \vec{B}, and why charged particles move in circles in magnetic fields.
• Scalar (Dot) Product — the complementary operation: same two vectors, but producing a scalar via \cos\theta instead of a vector via \sin\theta.
• Vectors: Introduction and Notation — the foundational article on what vectors are, how to add them, and how to decompose them into components.