Vector Algebra · Lesson 7 of 9

The Cross Product and Orientation

“Understand the magnitude and direction of a vector product using right-handed orientation.”

Learning Objectives

• Define the cross product and distinguish it from the dot product. • Use the right-hand rule to determine the perpendicular direction. • Apply basis-vector products and reverse their order correctly. • Interpret zero and maximum cross-product magnitudes. • Recognise the angle ambiguity in a sine-based calculation.

A product with a new direction

The dot product produces a scalar, but some geometric questions ask for a vector perpendicular to two given directions. The cross product supplies such a vector. Its magnitude depends on the two input lengths and their angle, while its orientation depends on the order of the inputs.

Definition

Vector or cross product

For two nonzero, nonparallel vectors, $a \times b$ is perpendicular to their plane. Its magnitude is $∣ a ∣∣ b ∣ sin θ$ , where $0 < θ < π$ , and its direction is chosen by the right-hand rule.

Cross-product definitionLaTeX

a \times b = ∣ a ∣ ∣ b ∣ sin θ \overset{n}{^}

\overset{n}{^}

is the unit normal chosen by turning from the first vector

a

towards the second vector

b

using right-handed orientation. Parallel inputs and zero inputs have zero cross product.

There are two opposite normal directions to a plane, so saying only “perpendicular to both” is incomplete. The orientation rule chooses one. When the product is zero, its direction is not defined; there is no need to select a unique normal in that case.

The right-hand rule

Hold the fingers of your right hand so that they curl from the first vector towards the second through the angle between them. Your thumb indicates the cross-product direction. A right-handed screw gives the same convention: turning it in that sense makes it advance along the chosen normal.

In a right-handed coordinate system, turning from the positive x-axis towards the positive y-axis selects the positive z-axis. This is the meaning of $\hat{i} \times \hat{j} = \hat{k}$ . The same orientation convention continues cyclically through the three basis directions.

Right-handed orientation— The x-to-y turn selects positive z. Reverse the turn and the normal reverses.

Example — A perpendicular direction

Problem
What is the direction of $(3 \hat{i}) \times (2 \hat{j})$ ?

1.The angle between $\hat{i}$ and $\hat{j}$ is $π /2$ , so the magnitude is $3 \times 2 \times 1 = 6$ .
2.Turning from positive x to positive y selects positive z by the right-hand rule.
3.Thus the product is $6 \hat{k}$ . The positive direction, not just the magnitude 6, is part of the answer.

The basis-vector cycle

The three forward cyclic products have positive signs. Reversing any pair changes the sign, and crossing any basis vector with itself gives zero. Learning this small system through orientation is more reliable than memorising unrelated products.

Forward and reverse basis productsLaTeX

\hat{i} \times \hat{j} \hat{j} \times \hat{i} \hat{i} \times \hat{i} = \hat{k}, = - \hat{k}, = 0, \hat{j} \times \hat{k} \hat{k} \times \hat{j} \hat{j} \times \hat{j} = \hat{i}, = - \hat{i}, = 0, \hat{k} \times \hat{i} \hat{i} \times \hat{k} \hat{k} \times \hat{k} = \hat{j} = - \hat{j} = 0

The cycle is

\hat{i} \to \hat{j} \to \hat{k} \to \hat{i}

. Along the cycle the product is positive; against it the product is negative.

The cyclic basis products— Following the arrows gives i × j = k, j × k = i, and k × i = j.

Example — A reversed basis pair

Problem
Evaluate $(2 \hat{j}) \times (3 \hat{i})$ and compare it with $(3 \hat{i}) \times (2 \hat{j})$ .

1.The first has magnitude 6, but its basis product is $\hat{j} \times \hat{i} = - \hat{k}$ . Thus it is $- 6 \hat{k}$ .
2.The reversed product is $6 \hat{k}$ .
3.The two magnitudes are equal and the directions are opposite. This demonstrates why the order of a cross product matters.

Reversing the orderLaTeX

a \times b = - b \times a

The cross product is anticommutative. Reversing the turn reverses the normal while leaving the length unchanged.

When the original product is zero, its negative is also zero, so the two orders happen to give the same result. Anticommutativity is the exact rule; it does not mean that every numerical comparison of the two orders must be unequal.

When does the product vanish?

The sine factor is zero at angles zero and 180 degrees. Nonzero vectors pointing along the same line therefore have zero cross product whether their directions agree or oppose. A zero input also forces a zero product, although no angle exists for that input.

Collinearity test for nonzero vectorsLaTeX

a \times b = 0 ⟺ a and b are collinear

This geometric statement assumes both inputs are nonzero. In particular,

a \times a = 0

and

a \times (- a) = 0

Example — A zero product without a zero input

Problem
Let $a = 2 \hat{i}$ and $b = - 5 \hat{i}$ . Find both their cross and dot products.

1.The vectors are nonzero and opposite, so their angle is $π$ . Hence $a \times b = 0$ .
2.The dot product is $2 (- 5) = - 10$ , which is not zero.
3.This is a counterexample to the claim that a zero cross product requires one of the vectors to be zero. It also shows why a zero cross product and a zero dot product have different meanings.

Magnitude, maximum value and angles

For fixed input lengths, the cross-product magnitude is largest when the vectors are perpendicular. It then equals the product of their lengths. If only the magnitude is known, the sine of their angle can be found, but sine usually does not distinguish an acute angle from its obtuse supplement.

Angle information from a cross productLaTeX

sin θ = \frac{∣ a \times b ∣}{∣ a ∣ ∣ b ∣}, ∣ a \times b ∣ \leq ∣ a ∣ ∣ b ∣

Both inputs must be nonzero. For fixed lengths, the maximum occurs at

θ = π /2

Example — Two possible angles

Problem
If $∣ a ∣ = 3$ , $∣ b ∣ = 2$ and $∣ a \times b ∣ = 3$ , what angles are possible?

1.The formula gives $sin θ = 3/ (3 \times 2) = 1/2$ .
2.In the interval $0 \leq θ \leq π$ , this means $θ = π /6$ or $5 π /6$ .
3.The sign of a dot product, or other direction information, is needed to choose between the acute and obtuse possibilities.

Example — Cross and dot products together

Problem
If $a \cdot b = 0$ and $a \times b = 0$ , what follows?

1.Assume for contradiction that both vectors are nonzero.
2.The dot product would require $θ = π /2$ , while the cross product would require $θ = 0$ or $π$ . These requirements cannot both hold.
3.Therefore at least one input vector is zero. This conclusion needs both equations; either equation on its own is insufficient.

Common mistake

The cross product is a vector, while the dot product is a scalar. A cross product gives a perpendicular direction only when it is nonzero. Do not select an angle from inverse sine without checking the supplementary possibility.

Quick check

Which is $\hat{j} \times \hat{k}$ ?

Quick check

Which is $\hat{i} \times \hat{k}$ ?

Quick check

For fixed nonzero magnitudes, when is cross-product magnitude greatest?

Quick check

Two nonzero vectors have zero cross product. They must be

Quick check

If $sin θ = 1/2$ for the angle between two vectors, the possibilities are

Practice Problems

Evaluate $\hat{k} \times \hat{i}$ , $\hat{k} \times \hat{j}$ and $(- 2 \hat{i}) \times (3 \hat{j})$ . Explain every sign.
Use a sketch or your right hand to explain why reversing a cross product reverses its direction.
If $∣ a ∣ = 4$ , $∣ b ∣ = 5$ and their angle is $π /6$ , find $∣ a \times b ∣$ .
Give two nonzero vectors whose cross product is zero. Give two nonzero vectors whose dot product is zero.
If $∣ a ∣ = 3$ , $∣ b ∣ = 2 /3$ and $a \times b$ is a unit vector, find all possible angles between the inputs.
Explain why the cross product of a vector with itself is $0$ , while its dot product with itself need not be zero.
Prove that simultaneous zero dot and cross products require at least one zero input.

Key Takeaways

A cross product is perpendicular to its inputs and has sine-based magnitude. The right-hand rule fixes which normal direction is selected. Reversing input order negates the cross product. Nonzero collinear inputs give zero cross product; perpendicular inputs maximise its magnitude. Cross-product magnitude alone may leave two supplementary angle possibilities.

Previous · Lesson 6

Projections and Dot Product Applications

Next · Lesson 8

Calculating Cross Products and Areas

A product with a new direction

The right-hand rule

The basis-vector cycle

When does the product vanish?

Magnitude, maximum value and angles

Quiz

Practice Problems

Key Takeaways