Cross product — formula, direction (right-hand rule), torque - area calculation: Worked examples

This page is a drill through every kind of cross-product problem you can meet. Before touching numbers, we lay out a map of all the cases, then work examples that visit each square on that map. If you have not yet met the magnitude rule $∣ A \times B ∣ = A B sin θ$ or the determinant formula, build them first in Cross product — formula, direction (right-hand rule), torque - area calculation.

Tools you'll use on every example

Before the drills, let's pin down the two pieces of notation that appear in every box below, so nothing is used before it is defined.

Formula The

$3 \times 3$ cofactor expansion (memorise this pattern) To cross $A = (A_{x}, A_{y}, A_{z})$ with $B = (B_{x}, B_{y}, B_{z})$ we expand a determinant along its top row: $A \times B = \hat{i} A_{x} B_{x} \hat{j} A_{y} B_{y} \hat{k} A_{z} B_{z} = + \hat{i} A_{y} B_{y} A_{z} B_{z} - \hat{j} A_{x} B_{x} A_{z} B_{z} + \hat{k} A_{x} B_{x} A_{y} B_{y}$ How to read it: for each of $\hat{i}, \hat{j}, \hat{k}$ you cover its own column, then cross-multiply the leftover $2 \times 2$ box as (top-left × bottom-right) − (top-right × bottom-left). Written out in full: $A \times B = (A_{y} B_{z} - A_{z} B_{y}) \hat{i} - (A_{x} B_{z} - A_{z} B_{x}) \hat{j} + (A_{x} B_{y} - A_{y} B_{x}) \hat{k}$ The signs are $+, -, +$ — the middle ( $\hat{j}$ ) term always carries a minus. That alternating sign is the "checkerboard" cofactor pattern from Determinants — 3×3 expansion; here it is what makes the cross product anti-commutative.

The scenario matrix

Every cross-product problem is really one of these cells. The columns tell you what changes; the rows tell you what kind of task.

Cell	Case class	What is special about it	Example
A	Perpendicular, in-plane ( $θ = 9 0^{\circ}$ )	$sin θ = 1$ , magnitude is maximum $= A B$	Ex 1
B	Parallel / antiparallel ( $θ = 0^{\circ}$ or $18 0^{\circ}$ )	$sin θ = 0$ , result is the zero vector	Ex 2
C	General 3-D components	all three output components nonzero	Ex 3
D	Sign / order flip (anti-commutativity)	$A \times B = - B \times A$	Ex 4
E	Geometric: triangle / parallelogram area	use $\tfrac12	\vec A\times\vec B
F	Angle-and-magnitude form (no components given)	use $A B sin θ$ directly	Ex 6
G	Real-world word problem (torque)	translate "push at a distance" into $r \times F$	Ex 7
H	Exam twist: solve for an unknown	work the cross product backwards	Ex 8
I	Limiting / degenerate: zero input	a zero vector kills the product	Ex 9

Example 1 — Cell A: perpendicular vectors (maximum case)

The figure below draws this case. The blue arrow is $A = 3 \hat{i}$ along $x$ , the orange arrow is $B = 4 \hat{j}$ along $y$ ; the small square marks their $9 0^{\circ}$ corner. The green rectangle is the parallelogram they span — because they are perpendicular it is a plain rectangle of area $3 \times 4 = 12$ , which is exactly $∣ A \times B ∣$ . The red dotted circle is the standard symbol for an arrow coming out of the page toward you: that is $A \times B = 12 \hat{k}$ . Notice how the maximum area and the out-of-page thumb both appear in one picture.

Figure — Cross product — formula, direction (right-hand rule), torque - area calculation

Example 2 — Cell B: parallel and antiparallel (the zero case)

Worked example When the arrows point the same way

(a) $A = 2 \hat{i} + \hat{j}$ , $B = 4 \hat{i} + 2 \hat{j}$ (parallel). (b) $C = 2 \hat{i} + \hat{j}$ , $D = - 6 \hat{i} - 3 \hat{j}$ (antiparallel). Find both cross products.

Forecast: what does a parallelogram with no width look like? What is its area?

(a) Notice $B = 2 A$ . They lie on the same line, $θ = 0^{\circ}$ . Why this step? Spotting the scalar multiple tells you they're parallel without computing an angle.
Compute the determinant. $A \times B = \hat{i} 24 \hat{j} 12 \hat{k} 00 = (1 \cdot 0 - 0 \cdot 2) \hat{i} - (0 - 0) \hat{j} + (2 \cdot 2 - 1 \cdot 4) \hat{k} = (4 - 4) \hat{k} = 0$ . Why this step? $sin 0^{\circ} = 0$ predicts a zero result; the components confirm it.
(b) $D = - 3 C$ , so $θ = 18 0^{\circ}$ . $sin 18 0^{\circ} = 0$ too, so $C \times D = 0$ . Why this step? Antiparallel is just $θ = 18 0^{\circ}$ ; $sin$ is zero at both $0^{\circ}$ and $18 0^{\circ}$ .

Verify: the parallelogram spanned by parallel vectors is a flat line — zero area — matching $0$ . Determinant of (b): $\hat{i} 2 - 6 \hat{j} 1 - 3 \hat{k} 00 = (2 \cdot - 3 - 1 \cdot - 6) \hat{k} = (- 6 + 6) \hat{k} = 0$ . ✓

Example 3 — Cell C: general 3-D vectors

Worked example All three components alive

$A = \hat{i} + 2 \hat{j} + 3 \hat{k}$ , $B = 4 \hat{i} + 5 \hat{j} + 6 \hat{k}$ . Find $A \times B$ .

Forecast: none of the components are zero, so expect a fully 3-D answer vector.

$\hat{i}$ component: $A_{y} B_{z} - A_{z} B_{y} = (2) (6) - (3) (5) = 12 - 15 = - 3$ . Why this step? Cover the $\hat{i}$ column and cross-multiply the remaining $2 \times 2$ — exactly the first term of the cofactor formula above.
$\hat{j}$ component: $- (A_{x} B_{z} - A_{z} B_{x}) = - [(1) (6) - (3) (4)] = - (6 - 12) = 6$ . Why this step? The middle term carries the built-in minus of the $+, -, +$ pattern.
$\hat{k}$ component: $A_{x} B_{y} - A_{y} B_{x} = (1) (5) - (2) (4) = 5 - 8 = - 3$ . Why this step? Last cofactor is positive again ( $+, -, +$ ).

$A \times B = - 3 \hat{i} + 6 \hat{j} - 3 \hat{k}$

Verify (perpendicularity test): the result must be perpendicular to both inputs, so each dot product should be zero. $A \cdot (A \times B) = (1) (- 3) + (2) (6) + (3) (- 3) = - 3 + 12 - 9 = 0$ ✓ and $B \cdot (A \times B) = (4) (- 3) + (5) (6) + (6) (- 3) = - 12 + 30 - 18 = 0$ ✓. (Same idea as the Dot product — formula, projection, work calculation.)

Example 4 — Cell D: order matters (anti-commutativity)

Worked example Swap the order, flip the thumb

Using $A, B$ from Example 3, compute $B \times A$ and compare.

Forecast: predict the answer before computing — the rule says it should be minus the previous one.

Recall the rule. $B \times A = - (A \times B)$ . Why this step? Curling right-hand fingers from $B$ to $A$ (instead of $A$ to $B$ ) puts the thumb on the opposite side.
Apply it. $- (- 3 \hat{i} + 6 \hat{j} - 3 \hat{k}) = 3 \hat{i} - 6 \hat{j} + 3 \hat{k}$ . Why this step? Same magnitude, reversed sense — negate every component.
Independent check by determinant. Swap the two rows of Example 3's matrix; swapping two rows of a determinant flips its sign (see Determinants — 3×3 expansion). Why this step? Confirms the negation is not a fluke but a determinant property.

Verify: $B \times A = 3 \hat{i} - 6 \hat{j} + 3 \hat{k} = - (A \times B)$ . Magnitudes equal, directions opposite. ✓

Example 5 — Cell E: area of a triangle in 3-D

Worked example A slanted triangle

Triangle with vertices $P (1, 0, 0)$ , $Q (0, 1, 0)$ , $R (0, 0, 1)$ . Find its area.

Forecast: this triangle is tilted in space — its area is not simply base×height in one plane. The cross product handles the tilt automatically.

Build two edge vectors from the same vertex $P$ . $P Q = Q - P = (- 1, 1, 0)$ , $P R = R - P = (- 1, 0, 1)$ . Why this step? The cross product needs both vectors sharing a tail; $P$ is the shared corner.
Cross them. $P Q \times P R = \hat{i} - 1 - 1 \hat{j} 10 \hat{k} 01 = (1 \cdot 1 - 0 \cdot 0) \hat{i} - ((- 1) (1) - 0 \cdot (- 1)) \hat{j} + ((- 1) (0) - 1 \cdot (- 1)) \hat{k} = \hat{i} + \hat{j} + \hat{k}$ . Why this step? The result vector is perpendicular to the tilted triangle — its length is the parallelogram area.
Halve the magnitude. $∣ P Q \times P R ∣ = 1^{2} + 1^{2} + 1^{2} = 3$ . Area $= \frac{1}{2} 3 \approx 0.866$ . Why this step? A triangle is half the parallelogram spanned by its two edges.

Verify: this is an equilateral triangle with side $2$ (e.g. $∣ P Q ∣ = (- 1)^{2} + 1^{2} = 2$ ). Equilateral area $= \frac{3}{4} s^{2} = \frac{3}{4} (2) = \frac{3}{2} = \frac{1}{2} 3$ . ✓

The figure below shows this tilted triangle in 3-D. The three labelled corners $P, Q, R$ sit one unit out along each axis; the green shaded face is the triangle itself, clearly not lying flat in any coordinate plane. The blue and orange arrows are the two edge vectors $P Q$ and $P R$ drawn from the shared corner $P$ . The red arrow springing from the centre is $P Q \times P R$ : notice it stands perpendicular to the green face — that perpendicularity is why its length measures the tilted area without us needing to un-tilt anything.

Example 6 — Cell F: angle-and-magnitude form (no components)

Worked example Only lengths and an angle given

Two vectors have magnitudes $A = 5$ and $B = 8$ with an angle $θ = 3 0^{\circ}$ between them. Find $∣ A \times B ∣$ and the parallelogram area.

Forecast: with no components, only the $A B sin θ$ form applies. Guess whether the answer is closer to $0$ or to $A B = 40$ .

Pick the right formula. We have magnitudes and an angle, not $x, y, z$ parts, so use $∣ A \times B ∣ = A B sin θ$ . Why this step? The determinant needs components we don't have; $A B sin θ$ needs exactly what we do have.
Substitute. $= 5 \cdot 8 \cdot sin 3 0^{\circ} = 40 \cdot \frac{1}{2} = 20$ . Why this step? $sin 3 0^{\circ} = \frac{1}{2}$ is a standard value.
State the geometric meaning. The parallelogram area is exactly this magnitude, $20$ square units. Why this step? $∣ A \times B ∣$ is the parallelogram area by definition.

Verify: $sin 3 0^{\circ} = 0.5$ , so $40 \times 0.5 = 20$ . Since $θ < 9 0^{\circ}$ , the answer is below the max $A B = 40$ , as expected. ✓

Example 7 — Cell G: real-world word problem (torque)

Worked example Pushing a wrench

A mechanic pushes a wrench with force $F = (0, - 20, 0) N$ (straight down) at the end of the handle located at $r = (0.3, 0, 0) m$ from the bolt. Find the torque $τ = r \times F$ .

Forecast: which way will the bolt turn — into the page or out of it? Guess before computing.

Translate words to vectors. "0.3 m along the handle" $= r = (0.3, 0, 0)$ ; "20 N downward" $= F = (0, - 20, 0)$ . Why this step? Torque needs both the position and the force as vectors from the pivot.
Cross them. $τ = \hat{i} 0.3 0 \hat{j} 0 - 20 \hat{k} 00 = (0 - 0) \hat{i} - (0 - 0) \hat{j} + ((0.3) (- 20) - 0) \hat{k} = - 6 \hat{k} N\cdotm$ . Why this step? $r$ along $+ x$ , $F$ along $- y$ , and $\hat{i} \times (- \hat{j}) = - \hat{k}$ .
Interpret the sign. $- \hat{k}$ points into the page → clockwise turning. Why this step? The direction of $τ$ is the axis about which the bolt tends to rotate (see Torque and rotational equilibrium).

Verify: magnitude $∣ τ ∣ = r F sin 9 0^{\circ} = (0.3) (20) (1) = 6 N\cdotm$ , matching $∣ - 6 \hat{k} ∣ = 6$ . Units: m × N = N·m ✓.

Example 8 — Cell H: exam twist — solve backwards

Worked example Find the missing component

$A = (2, 0, 1)$ and $B = (0, 3, b)$ . Given that the $\hat{i}$ -component of $A \times B$ equals $- 3$ , find $b$ .

Forecast: you must run the cross-product machinery in reverse — set the formula equal to the given number and solve.

Write the $\hat{i}$ -component formula. $(A \times B)_{x} = A_{y} B_{z} - A_{z} B_{y} = (0) (b) - (1) (3) = - 3$ . Why this step? Only the $\hat{i}$ -component is constrained, so only that formula is needed.
Notice it is already fixed. $(0) (b) - 3 = - 3$ holds for every $b$ — the constraint is automatically satisfied. Why this step? $A_{y} = 0$ makes the $b$ -term vanish, so $b$ is unconstrained by the $\hat{i}$ -equation. The "twist" is that the answer is "any $b$ ".
Sanity via a different component. If instead they had constrained $\hat{j}$ : $(A \times B)_{y} = - (A_{x} B_{z} - A_{z} B_{x}) = - (2 b - 0) = - 2 b$ , which does pin down $b$ . This shows which equation carries the information. Why this step? Exam twists often hide the fact that the chosen component is degenerate — always test another.

Verify: plug $b = 5$ : $\hat{i}$ -comp $= (0) (5) - (1) (3) = - 3$ ✓; plug $b = - 1$ : $(0) (- 1) - 3 = - 3$ ✓. Both work — confirming any $b$ satisfies it.

Example 9 — Cell I: degenerate zero input

Recall Quick self-test (click to expand)

This is a fold-out flashcard block: each line states a question, and everything after the ::: is the hidden answer — cover it, answer from memory, then reveal.

Which cell has $sin θ = 1$ ? ::: Cell A (perpendicular, $θ = 9 0^{\circ}$ ) Which two cells give the zero vector? ::: Cell B (parallel/antiparallel) and Cell I (zero input) To get triangle area you multiply the magnitude by what? ::: $\frac{1}{2}$ How do you check a computed $A \times B$ is correct? ::: Dot it with $A$ and $B$ ; both must be $0$ (perpendicularity) If you swap the order of the vectors, the result... ::: flips sign (anti-commutative)

Connections

Cross product — formula, direction (right-hand rule), torque - area calculation (parent — the theory these examples drill)
Dot product — formula, projection, work calculation (used as the perpendicularity check)
Determinants — 3×3 expansion (the cofactor pattern and the row-swap sign flip)
Vectors — addition, components, unit vectors (defining $\hat{i}, \hat{j}, \hat{k}$ and building edge vectors from points)
Torque and rotational equilibrium (Example 7)
Angular momentum — L = r × p (same $r \times (\dots)$ structure)
Magnetic force — F = qv × B (another physical cross product)