Computing Cross Products and Determining Direction with the Right-Hand Rule

Welcome back. In the previous lesson, the dot product combined two vectors into a scalar that measures directional alignment. The cross product does almost the complementary job: it combines two vectors into a new vector that records their perpendicular separation and a sense of rotation.

This completes the vector toolkit needed before calculus. You will learn to find a cross product from magnitudes and angle or from Cartesian components, and—just as importantly—to determine its direction consistently with the right-hand rule. These ideas will return when we study torque, angular momentum, and magnetic forces.

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What a cross product represents

For two vectors $\vec{A}$ and $\vec{B}$, their cross product is written

$$\vec{A}\times \vec{B}.$$

Unlike the dot product, its result is a vector, not a number. Let

$$\vec{C}=\vec{A}\times \vec{B}.$$

The vector $\vec{C}$ has two defining features:

1. It is perpendicular to both $\vec{A}$ and $\vec{B}$.
2. Its magnitude is

$$\boxed{|\vec{A}\times \vec{B}|=|\vec{A}||\vec{B}|\sin \theta }$$

where $\theta$ is the angle between the vectors, taken from $0$ to $\pi$.

Although we may draw $\vec{A}$ and $\vec{B}$ on a flat page, their cross product points in a third direction: either out of the page or into it. Thus, the usual vector cross product is a three-dimensional operation.

The sine in the magnitude formula is significant. Recall that $\sin \theta$ is zero when the vectors are parallel or antiparallel, and it is largest when they are perpendicular. Therefore:

Relationship of $\vec{A}$ and $\vec{B}$	Cross-product magnitude
Parallel or antiparallel	$0$
At an intermediate angle	Between $0$ and its maximum
Perpendicular	$|\vec{A}||\vec{B}|$, its maximum

A useful geometric interpretation follows. If $\vec{A}$ and $\vec{B}$ form adjacent sides of a parallelogram, then

$$\boxed{\text{area of parallelogram}=|\vec{A}\times \vec{B}|}$$

This works because the base can be $|\vec{A}|$, while the height is the perpendicular part of $\vec{B}$:

$$\text{height}=|\vec{B}|\sin \theta .$$

So the cross product measures not alignment, as the dot product does, but the extent to which two vectors span an area.

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Direction: the right-hand rule and order

There are always two directions perpendicular to the plane containing $\vec{A}$ and $\vec{B}$. The right-hand rule chooses one of them.

Place the vectors tail-to-tail. Then use your right hand in either of these equivalent ways:

• Point the fingers of your right hand along the first vector, $\vec{A}$, and curl them toward the second vector, $\vec{B}$. Your extended thumb gives the direction of $\vec{A}\times \vec{B}$.
• Point your index finger along $\vec{A}$, your middle finger along $\vec{B}$, and your thumb then points along $\vec{A}\times \vec{B}$.

The order is not a cosmetic detail. Reversing the order reverses the direction:

$$\boxed{\vec{B}\times \vec{A}=-\left(\vec{A}\times \vec{B}\right)}$$

This is called anticommutativity. In particular,

$$\vec{A}\times \vec{B}\ne \vec{B}\times \vec{A}$$

unless the cross product is zero.

For vectors drawn in the $xy$-plane, use the standard three-dimensional convention:

• positive $x$: right
• positive $y$: up
• positive $z$: out of the page
• negative $z$: into the page

For example, let $\vec{A}$ point right and $\vec{B}$ point up. Curling from right toward up is counterclockwise, so

$$\vec{A}\times \vec{B}$$

points out of the page, in the positive $z$-direction. If you reverse the two vectors, the result points into the page.

The standard unit-vector results summarize this convention:

$$\hat{ı}\times \hat{ȷ}=\hat{k},$$ $$\hat{ȷ}\times \hat{k}=\hat{ı},$$ $$\hat{k}\times \hat{ı}=\hat{ȷ}.$$

The cyclic order

$$\hat{ı},\hat{ȷ},\hat{k}$$

gives a positive result. Reversing any pair introduces a minus sign:

$$\hat{ȷ}\times \hat{ı}=-\hat{k}.$$

A vector crossed with itself is zero:

$$\vec{A}\times \vec{A}=\vec{0}.$$

That fits the magnitude formula because the angle from a vector to itself is zero, and $\sin 0=0$.

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Computing a cross product from magnitudes and angle

When a problem gives two vector magnitudes and their included angle, use

$$|\vec{A}\times \vec{B}|=|\vec{A}||\vec{B}|\sin \theta .$$

Then use the right-hand rule to supply the direction.

Suppose $|\vec{A}|=6$, $|\vec{B}|=5$, and the angle from $\vec{A}$ to $\vec{B}$ is $30^{\circ }$, counterclockwise in the $xy$-plane. The magnitude is

$$|\vec{A}\times \vec{B}|=(6)(5)\sin 30^{\circ }.$$

Since

$$\sin 30^{\circ }=\frac{1}{2},$$

we obtain

$$|\vec{A}\times \vec{B}|=15.$$

Because the turn from the first vector to the second is counterclockwise in the $xy$-plane, the right-hand rule gives positive $z$. Thus,

$$\boxed{\vec{A}\times \vec{B}=15\hat{k}}$$

if the vectors have compatible units or are unitless.

Be alert to the distinction between magnitude and vector. The expression

$$|\vec{A}\times \vec{B}|$$

is a nonnegative scalar. The expression

$$\vec{A}\times \vec{B}$$

is a vector, so it includes direction.

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Computing a cross product from components

For three-dimensional vectors,

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

the component formula is

$$\boxed{\vec{A}\times \vec{B}=⟨A_y{B}_z-A_z{B}_y,A_z{B}_x-A_x{B}_z,A_x{B}_y-A_y{B}_x⟩}$$

There are two habits that make this formula much less error-prone:

1. Preserve the order exactly as written: first $\vec{A}$, then $\vec{B}$.
2. Keep each component grouped in parentheses until the final subtraction is complete.

Consider

$$\vec{A}=⟨2,-1,3⟩,\vec{B}=⟨1,4,-2⟩.$$

The $x$-component is

$$A_y{B}_z-A_z{B}_y=(-1)(-2)-(3)(4)=2-12=-10.$$

The $y$-component is

$$A_z{B}_x-A_x{B}_z=(3)(1)-(2)(-2)=3+4=7.$$

The $z$-component is

$$A_x{B}_y-A_y{B}_x=(2)(4)-(-1)(1)=8+1=9.$$

Therefore,

$$\boxed{\vec{A}\times \vec{B}=⟨-10,7,9⟩}$$

A built-in check uses the dot product from the previous lesson. The cross product must be perpendicular to both original vectors, so its dot product with each should be zero:

$$\vec{A}\cdot \left(\vec{A}\times \vec{B}\right)=2(-10)+(-1)(7)+3(9)=0,$$ $$\vec{B}\cdot \left(\vec{A}\times \vec{B}\right)=1(-10)+4(7)+(-2)(9)=0.$$

That does not replace careful arithmetic, but it is an excellent way to catch a sign error.

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The important two-dimensional case

Much early mechanics is drawn in two dimensions, but the cross product still has a three-dimensional result. Treat a two-dimensional vector as having a zero $z$-component:

$$\vec{A}=⟨A_x,A_y,0⟩,\vec{B}=⟨B_x,B_y,0⟩.$$

Substituting into the general component formula gives

$$\boxed{\vec{A}\times \vec{B}=⟨0,0,A_x{B}_y-A_y{B}_x⟩}$$

The result is always along the $z$-axis:

• A positive $z$-component means out of the page.
• A negative $z$-component means into the page.
• A zero result means the vectors are parallel, antiparallel, or one is the zero vector.

For instance, let

$$\vec{r}=⟨0.30,0,0⟩\mathrm{m}$$

be a position vector from a pivot to the point where a force is applied, and let

$$\vec{F}=⟨0,50,0⟩\mathrm{N}.$$

Then

$$\vec{r}\times \vec{F}=⟨0,0,(0.30)(50)-(0)(0)⟩.$$

So,

$$\boxed{\vec{r}\times \vec{F}=⟨0,0,15⟩\mathrm{N}\mathrm{m}}$$

The positive $z$-direction means out of the page. Physically, this vector describes the tendency of the force to rotate the object counterclockwise around the pivot. In the rotation module, this quantity will be named torque:

$$\vec{\tau }=\vec{r}\times \vec{F}.$$

The units of a cross product are the product of the input units. In this example, the units are $\mathrm{N}\mathrm{m}$. Torque shares these base units with energy, but torque is not energy: its vector nature and physical role are different.

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A compact decision guide

Use the information given in the problem to choose your method.

Given	Method
Magnitudes and angle between vectors	Find $|\vec{A}||\vec{B}|\sin \theta$, then apply the right-hand rule
Three-dimensional Cartesian components	Use the three-component formula
Two-dimensional Cartesian components	Use $⟨0,0,A_x{B}_y-A_y{B}_x⟩$
A result seems uncertain	Check that the result has zero dot product with both original vectors

The most common mistakes are:

• Using cosine instead of sine. Cross products involve the perpendicular component, so they use sine.
• Returning only a magnitude when the question requires the vector.
• Forgetting that $\vec{A}\times \vec{B}$ and $\vec{B}\times \vec{A}$ have opposite directions.
• Applying the right-hand rule from the second vector toward the first.
• Treating vectors in the plane as if the result also had to lie in that plane.

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Wrap-up

The cross product of two vectors is a vector perpendicular to both inputs:

$$|\vec{A}\times \vec{B}|=|\vec{A}||\vec{B}|\sin \theta .$$

Its magnitude is the area of the parallelogram spanned by the vectors, so it vanishes for parallel vectors and is largest for perpendicular vectors. Its direction comes from the right-hand rule and depends critically on the order of the vectors:

$$\vec{B}\times \vec{A}=-\left(\vec{A}\times \vec{B}\right).$$

From components, compute each coordinate by taking the appropriate difference of products; in two dimensions, the result lies along the positive or negative $z$-axis.

You have now completed the vector portion of this module. The next module begins calculus systematically with average rates of change, the numerical and graphical idea that leads naturally to limits and derivatives.