Vector (cross) product

Push open a door and something curious happens. Push straight at the hinge and nothing turns, no matter how hard you shove. Push at the outer edge, at right angles to the door, and it swings open with the lightest touch. The same force, applied at the same point, produces wildly different turning effects depending on the angle at which it acts. Ordinary multiplication of two vectors — the dot product — cannot capture this, because it only cares about how much two vectors point along each other. What the door needs is a rule that cares about how much two vectors point across each other, and hands back an answer that itself has a direction: the axis about which the turning happens. That rule is the vector product, or cross product, written $\vec A\times\vec B$ . It takes two vectors and manufactures a third one, perpendicular to both, whose length measures how "crossed" the originals are and whose direction is fixed by a right-hand convention every physics student has to internalise by feel, not just by formula. It was introduced briefly in Motion in a Plane; here it returns as the single mathematical gate through which the whole chapter "System of Particles and Rotational Motion" passes. Torque $\vec\tau=\vec r\times\vec F$ — the thing that actually opens the door — and angular momentum $\vec L=\vec r\times\vec p$ — the rotational analogue of ordinary momentum — are both cross products in disguise. Get comfortable with the magnitude rule $AB\sin\theta$ , the right-hand rule for direction, and the component formula, and torque and angular momentum become almost free. This article treats the cross product as a tool: how to compute it, why $\sin\theta$ rather than $\cos\theta$ , why order matters so much that swapping it flips the sign, and how the two worked examples from earlier study extend cleanly into the torque and angular-momentum language used for the rest of the chapter.

Key takeaways

The vector product of $\vec A$ and $\vec B$ is $\vec A\times\vec B = AB\sin\theta\,\hat n$ , a new vector perpendicular to both, with $\hat n$ fixed by the right-hand rule.
Order matters: the product is anti-commutative, $\vec A\times\vec B = -(\vec B\times\vec A)$ — swap the two vectors and the direction reverses.
The magnitude $AB\sin\theta$ is the area of the parallelogram spanned by $\vec A$ and $\vec B$ ; it is zero for parallel/anti-parallel vectors and maximum ( $=AB$ ) for perpendicular ones.
In component form, $\vec A\times\vec B = (A_yB_z-A_zB_y)\hat i + (A_zB_x-A_xB_z)\hat j + (A_xB_y-A_yB_x)\hat k$ ; for planar vectors only the $\hat k$ term survives.
Torque $\vec\tau = \vec r\times\vec F$ and angular momentum $\vec L = \vec r\times\vec p$ are the two cross products this chapter is built on — everything about "which way does it turn" traces back to this one operation.
The dot product ( $AB\cos\theta$ , a scalar) and the cross product ( $AB\sin\theta\,\hat n$ , a vector) are complementary: one is greatest when vectors are parallel, the other when they are perpendicular.
Key idea: the cross product $\vec A\times\vec B = AB\sin\theta\,\hat n$ manufactures a new vector perpendicular to both originals, with magnitude equal to the parallelogram area they span and direction fixed by the right-hand rule, and it is exactly this operation that defines torque and angular momentum for the rest of the chapter.

The core insight

The vector product $\vec A\times\vec B = AB\sin\theta\,\hat n$ turns two vectors into a third, perpendicular to both and fixed in sign by the right-hand rule, and this single operation is the engine behind torque and angular momentum in rotational motion.

Where it fits in the syllabus

The definition and geometry of the cross product were first met in Motion in a Plane; NCERT does not re-derive it inside "System of Particles and Rotational Motion," but the chapter leans on it immediately and constantly — the very first new quantities it defines, torque $\vec\tau=\vec r\times\vec F$ and angular momentum $\vec L=\vec r\times\vec p$ , are cross products by definition, and later, moment of inertia problems and equilibrium conditions all inherit the sign conventions fixed here. This article is therefore a prerequisite refresher, not a new syllabus item in its own right: everything below — magnitude, direction, component formula, anti-commutativity — is examinable insofar as it is needed to compute $\vec\tau$ and $\vec L$ correctly, and no further vector-algebra machinery (triple products, general 3-D geometry proofs) is required at this level.

The physical picture

Two vectors $\vec A$ and $\vec B$ , drawn from a common point, generally do not point the same way; they enclose some angle $\theta$ between $0°$ and $180°$ . The cross product is defined to capture exactly the "sideways," turning-sensitive part of that relationship, as a new vector:

\vec A\times\vec B = AB\sin\theta\,\hat n

where: $A=|\vec A|$ and $B=|\vec B|$ are the magnitudes of the two vectors, $\theta$ is the angle between them measured the short way (so $0°\le\theta\le180°$ , and $\sin\theta$ is never negative), and $\hat n$ is a unit vector perpendicular to the plane containing $\vec A$ and $\vec B$ , whose sense is fixed by the right-hand rule: point the fingers of the right hand along $\vec A$ , curl them toward $\vec B$ through the smaller angle, and the thumb gives $\hat n$ . The physical reading is that $|\vec A\times\vec B|=AB\sin\theta$ is exactly the area of the parallelogram with sides $\vec A$ and $\vec B$ — a geometric fact worth remembering, because it makes the $\sin\theta$ dependence visual rather than something to memorise blindly. When $\vec A$ and $\vec B$ point the same or opposite way ( $\theta=0°$ or $180°$ ) the "parallelogram" collapses to a line of zero area, so the cross product vanishes even though neither vector is zero; when they are perpendicular ( $\theta=90°$ ) the parallelogram is a full rectangle of area $AB$ , the largest possible value.

Figure

Two vectors A and B drawn from a common origin with the angle theta between them, the parallelogram they span shaded to show its area equals A B sine theta, and a right hand shown curling from A toward B with the thumb pointing straight up out of the page along n hat

Working with vectors written out in $\hat i,\hat j,\hat k$ components is faster and avoids ever having to see the parallelogram. With $\vec A = A_x\hat i + A_y\hat j + A_z\hat k$ and $\vec B = B_x\hat i + B_y\hat j + B_z\hat k$ , the cross product expands to

\vec A\times\vec B = (A_yB_z - A_zB_y)\hat i + (A_zB_x - A_xB_z)\hat j + (A_xB_y - A_yB_x)\hat k

where: each bracket pairs the "other two" components of $\vec A$ and $\vec B$ in a fixed cyclic pattern ( $x\!\to\! y\!\to\! z\!\to\! x$ ), and the physical reading is that this formula is just the magnitude-and-direction definition rewritten in coordinates — it always agrees with $AB\sin\theta\,\hat n$ , but never requires you to measure an angle. Most problems in this chapter keep position vectors and forces in the x-y plane, so $A_z=B_z=0$ and the first two brackets vanish automatically, leaving the single useful line

\vec A\times\vec B = (A_xB_y - A_yB_x)\hat k

which is the version actually used to compute torque and angular momentum for planar problems throughout the chapter.

Derivation AWhy the cross product is anti-commutative

The goal here is to show, starting only from the definition $\vec A\times\vec B = AB\sin\theta\,\hat n$ and the geometric meaning of the right-hand rule, that reversing the order of the two vectors — computing $\vec B\times\vec A$ instead of $\vec A\times\vec B$ — must reverse the sign of the answer, $\vec B\times\vec A = -(\vec A\times\vec B)$ , rather than simply asserting it as a rule to memorise. This matters because the entire sign convention used for torque and angular momentum later in the chapter rests on taking $\vec r\times\vec F$ and $\vec r\times\vec p$ in that exact order.

Step 1. Compare magnitudes first. The magnitude of $\vec A\times\vec B$ is $AB\sin\theta$ , where $\theta$ is the angle between $\vec A$ and $\vec B$ . Computing $\vec B\times\vec A$ instead swaps which vector is "first," but the angle between the same pair of vectors, and hence $\sin\theta$ , is unchanged — angle is a property of the pair, not of the order in which you name them. So $|\vec B\times\vec A| = BA\sin\theta = AB\sin\theta = |\vec A\times\vec B|$ : the two products have identical magnitude, and any difference must live entirely in the direction.

Step 2. Track the direction using the right-hand rule. For $\vec A\times\vec B$ , curl the fingers of the right hand starting from $\vec A$ and sweeping toward $\vec B$ through the smaller angle; the thumb gives the direction, call it $\hat n$ . For $\vec B\times\vec A$ , the rule now says to start from $\vec B$ and sweep toward $\vec A$ instead — the opposite rotational sense around the same pair of vectors. Curling the same right hand the opposite way around the same axis flips which way the thumb points: it now points along $-\hat n$ .

Step 3. Combine the two results. Since $\vec B\times\vec A$ has the same magnitude $AB\sin\theta$ as $\vec A\times\vec B$ but the opposite direction $-\hat n$ instead of $\hat n$ , it follows that $\vec B\times\vec A = AB\sin\theta\,(-\hat n) = -(AB\sin\theta\,\hat n) = -(\vec A\times\vec B)$ . Reversing the order of the two vectors reverses the sign of the cross product, and no other change occurs — this is exactly the anti-commutativity property used, without further proof, whenever torque or angular momentum is written as $\vec r\times\vec F$ or $\vec r\times\vec p$ and never the other way round.

Worked examples

Board level - cross product of two planar vectors, and its physical reading as a torque

Find $\vec A\times\vec B$ if $\vec A = 3\hat i + 2\hat j$ (m) and $\vec B = 1\hat i + 4\hat j$ (N), and state the magnitude and direction of the result, reading $\vec A$ as a position vector and $\vec B$ as a force so the answer is a torque.

Step 1. Both vectors lie in the x-y plane ( $A_z=B_z=0$ ), so only the $\hat k$ term of the general component formula survives: $\vec A\times\vec B = (A_xB_y - A_yB_x)\hat k$ .

Step 2. Substitute the numbers: $A_xB_y - A_yB_x = (3)(4) - (2)(1) = 12 - 2 = 10$ .

Step 3. State the result with units. Since $\vec A$ carries units of metres and $\vec B$ units of newtons, the product carries units of $\text{N}\,\cdot\,\text{m}$ , the unit of torque: $\vec A\times\vec B = 10\,\hat k\ \text{N}\,\cdot\,\text{m}$ .

Step 4. Read off magnitude and direction. The magnitude is $10\ \text{N}\,\cdot\,\text{m}$ , and the direction is $+\hat k$ — by the right-hand rule, curling the fingers from $\vec A$ toward $\vec B$ sends the thumb out of the page, so the "turning effect" this pair would produce, taken as a torque, is anticlockwise when viewed from the $+\hat k$ side. The cross product is $\vec A\times\vec B = 10\,\hat k\ \text{N}\,\cdot\,\text{m}$ , magnitude $10\ \text{N}\,\cdot\,\text{m}$ , directed out of the page.

Competitive level - comparing the cross product and dot product at a fixed angle

Two vectors have magnitudes $A = 5$ and $B = 8$ (SI units), with an angle of $30°$ between them. Find the magnitude of $\vec A\times\vec B$ , compare it with $\vec A\cdot\vec B$ , and state at what angle the two magnitudes would be equal.

Step 1. Compute the cross-product magnitude directly from the definition: $|\vec A\times\vec B| = AB\sin\theta = 5\times8\times\sin30° = 40\times0.5 = 20$ .

Step 2. Compute the dot product for comparison: $\vec A\cdot\vec B = AB\cos\theta = 40\times\cos30° = 40\times0.866 \approx 34.6$ . At $\theta=30°$ the vectors are closer to parallel than perpendicular, so the dot product ( $\approx34.6$ ) dominates over the cross product ( $20$ ) — exactly as expected, since $\cos\theta$ is large and $\sin\theta$ is small for a small angle.

Step 3. Find the crossover angle. The two magnitudes $AB\sin\theta$ and $AB\cos\theta$ are equal when $\sin\theta=\cos\theta$ , i.e. $\tan\theta=1$ , so $\theta = 45°$ . Below $45°$ the dot product is larger; above $45°$ the cross product takes over, reaching its own maximum at $\theta=90°$ (where the dot product falls to zero). At $30°$ : $|\vec A\times\vec B| = 20$ and $\vec A\cdot\vec B \approx 34.6$ ; the two would be equal at $\theta = 45°$ . Timing note: this whole comparison is two substitutions and one small-angle table lookup — under a minute if $\sin30°=0.5$ , $\cos30°\approx0.866$ and $\tan\theta=1\Rightarrow\theta=45°$ are known cold; do not waste time re-deriving these standard angle values in an exam.

Common traps

Forgetting the cross product is anti-commutative: $\vec A\times\vec B = -(\vec B\times\vec A)$ , so silently swapping the order of $\vec r$ and $\vec F$ (or $\vec r$ and $\vec p$ ) later in the chapter flips the sign of torque or angular momentum — an easy way to lose a sign mark even when the magnitude is right.
Using $\cos\theta$ where $\sin\theta$ was needed, or vice versa — a fast way to confuse the cross product with the dot product; a quick check is that the cross product is a vector (needs a direction stated) while the dot product is a bare scalar.
Applying the right-hand rule in the wrong rotational sense — always curl from the first-named vector toward the second, never the other way, and always through the smaller angle.
Forgetting that $\vec A\times\vec B = \vec 0$ whenever the two vectors are parallel or anti-parallel, even though neither vector is individually zero — this is exactly why a force pushing straight along the line to the pivot produces zero torque.
Mixing up which component formula applies — dropping the $\hat i$ and $\hat j$ terms is only valid when both vectors genuinely have zero $z$ -component; skipping this check on a 3-D problem gives an incomplete answer.

Common trapThe cross product is not "just multiplication" — order and the perpendicular direction are both non-negotiable

The single biggest mark-loser is treating $\vec A\times\vec B$ like ordinary scalar multiplication, where $ab=ba$ and the answer is just a number. Neither is true here: the result is a new vector, perpendicular to the original plane, and reversing the input order reverses its sign. This is not a technicality — it is the entire reason torque and angular momentum carry a well-defined sense of rotation (clockwise versus anticlockwise, or equivalently into the page versus out of it). Write $\vec\tau=\vec r\times\vec F$ in that order, every time, and always finish a cross-product answer with both a magnitude and a stated direction (component form, or "out of the page" / "into the page") — a magnitude alone is an incomplete answer to any cross-product question.

You should be able to…

State the definition $\vec A\times\vec B = AB\sin\theta\,\hat n$ and correctly determine the direction of $\hat n$ using the right-hand rule.
Compute the cross product of two vectors given in $\hat i,\hat j,\hat k$ component form, including the simplified planar case where only the $\hat k$ term survives.
Explain, from the definition, why the cross product is zero for parallel or anti-parallel vectors and maximum for perpendicular vectors.
State and justify the anti-commutativity property $\vec A\times\vec B = -(\vec B\times\vec A)$ .
Compare the cross product ( $AB\sin\theta$ , a vector) with the dot product ( $AB\cos\theta$ , a scalar) at a given angle.
Recognise the cross product as the exact operation used to define torque $\vec\tau=\vec r\times\vec F$ and angular momentum $\vec L=\vec r\times\vec p$ later in the chapter.

Ask the tutor: if the right-hand rule direction or the anti-commutativity sign flip still feels shaky, ask the AI tutor to walk through two or three more component examples before moving on to torque.

Why does the cross product need sine of theta instead of just multiplying the magnitudes directly?

Because the cross product is built to measure how much two vectors point across each other, not along each other. Picture pushing a door: a force applied exactly along the line to the hinge produces zero turning effect, no matter how large the force, because there is nothing "sideways" about it — and indeed $\theta=0°$ or $180°$ gives $\sin\theta=0$ . A force applied at right angles to the door produces the maximum turning effect for a given force and distance, matching $\sin90°=1$ , the largest value sine can take. The sine function is exactly the mathematical shape of "how perpendicular are these two vectors," which is precisely what a turning effect depends on — so it is not an arbitrary choice, it is forced by the physics the cross product is meant to describe.

How do I remember which way the right-hand rule points without getting confused?

Point the four fingers of your right hand along the first vector in the product, $\vec A$ , then curl them, like closing a fist, toward the second vector, $\vec B$ , sweeping through the smaller of the two possible angles. Your extended thumb then points along $\hat n$ , the direction of $\vec A\times\vec B$ . The order really is "first vector, then curl toward second vector" — practising this physically with your own hand on a few examples (rather than only reading component formulas) is the fastest way to make it automatic, especially since this exact motion recurs for both torque and angular momentum later in the chapter.

Is the cross product only defined in three dimensions, and does that matter for this chapter?

Yes — the cross product as defined here (a new vector perpendicular to both inputs) is a genuinely three-dimensional construction; in two dimensions there is no third direction for the result to point along, so strictly the operation needs $\hat i,\hat j,\hat k$ space. In practice, almost every problem in this chapter keeps position vectors and forces confined to a flat page (the x-y plane), which is why the reduced formula $\vec A\times\vec B=(A_xB_y-A_yB_x)\hat k$ is so useful: the result always points purely along $\hat k$ , either out of the page or into it, and you never need to leave the plane you are working in to describe torque or angular momentum correctly.

Once I compute the cross product for torque, what do I actually do with the direction I get?

The direction of $\vec\tau=\vec r\times\vec F$ tells you the sense of rotation the force would produce about the chosen origin or pivot: by convention, $\vec\tau$ pointing out of the page (along $+\hat k$ ) corresponds to an anticlockwise turning tendency, and into the page ( $-\hat k$ ) to a clockwise one, exactly matching how you would read angular velocity and angular momentum vectors later in the chapter. You are not expected to do anything further with the direction at this stage beyond stating it correctly and, in equilibrium problems, adding torques as signed numbers (anticlockwise positive, clockwise negative, or vice versa, as long as you are consistent) — the full three-dimensional vector addition of torques is not required at Class 11 level.