What is the vector product of two vectors?

The vector (cross) product of two vectors A and B is a third vector C = A×B with magnitude AB sinθ, where θ is the angle between them, and direction perpendicular to the plane containing A and B, fixed by the right-hand rule. Because the result is itself a vector, the cross product is the natural tool for rotational quantities that have a direction along an axis — moment of a force (torque) and angular momentum are both defined as cross products.

Is the cross product commutative?

No. The vector product is anti-commutative: A×B = −(B×A). The two results have the same magnitude AB sinθ and both are perpendicular to the plane of A and B, but they point to opposite sides. This is why torque is written τ = r×F and never F×r — the order fixes the sense of rotation. The scalar (dot) product, by contrast, is commutative.

Why is A×A equal to zero?

Because the angle between a vector and itself is 0°, and sin 0° = 0, so the magnitude A·A·sin0° = 0. The cross product of any two parallel (or anti-parallel) vectors is the null vector. The same reasoning gives î×î = ĵ×ĵ = k̂×k̂ = 0. Whenever two vectors line up, their cross product vanishes.

What are the unit-vector cross products î×ĵ, ĵ×k̂, k̂×î?

For a right-handed coordinate system they follow the cyclic rule î×ĵ = k̂, ĵ×k̂ = î, k̂×î = ĵ. Going forward through the cycle î → ĵ → k̂ → î gives a positive result; reversing the order introduces a minus sign, so ĵ×î = −k̂, k̂×ĵ = −î, î×k̂ = −ĵ. Each product is a unit vector because the unit vectors are mutually perpendicular and sin 90° = 1.

How do I compute a cross product from components?

Write the 3×3 determinant with î, ĵ, k̂ in the top row, the components of the first vector in the second row, and the second vector in the third row. Expanding gives A×B = (A_yB_z − A_zB_y)î − (A_xB_z − A_zB_x)ĵ + (A_xB_y − A_yB_x)k̂. Note the minus sign on the middle (ĵ) term — it is the most common arithmetic slip in NEET cross-product problems.

What is the geometric meaning of the magnitude of A×B?

|A×B| = AB sinθ equals the area of the parallelogram whose two adjacent sides are A and B. Half of it, ½|A×B|, is the area of the triangle spanned by the two vectors. This is the fastest way to find the area of a triangle or parallelogram from coordinates: form the two side vectors and take half the magnitude of their cross product.

Why are torque and angular momentum defined as cross products?

Both quantities have a direction along the axis of rotation, not along either of the vectors that build them. Torque τ = r×F points along the axis about which the force tends to rotate the body; angular momentum L = r×p points along the axis of the circulation of momentum. Only the cross product produces a vector perpendicular to the plane of motion, so it is the only product that can carry this axial direction. Their magnitudes also vanish when r and F (or p) are parallel — exactly when there is no turning effect.

Vector Product of Two Vectors (Cross Product) — NEET Physics

Definition of the vector product

In the chapter on work and energy you met the scalar (dot) product, which combines two vectors into a single number — work is force "dotted" with displacement. NCERT §6.5 now defines a second product, one that combines two vectors into a third vector. Following NCERT, the vector product of two vectors $\vec a$ and $\vec b$ is a vector $\vec c = \vec a\times\vec b$ specified by three statements:

Its magnitude is $c = ab\sin\theta$, where $a$ and $b$ are the magnitudes of $\vec a$ and $\vec b$ and $\theta$ is the angle between them.
Its direction is perpendicular to the plane containing $\vec a$ and $\vec b$.
The sense along that perpendicular is fixed by the right-handed screw (or right-hand) rule, turning from $\vec a$ towards $\vec b$.

Compactly,

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

where $\hat n$ is the unit vector along that perpendicular direction. Because of the cross symbol, the vector product is also called the cross product. The factor $\sin\theta$ is the heart of its behaviour: it is zero when the vectors are parallel ($\theta = 0^\circ$) or anti-parallel ($\theta = 180^\circ$) and maximum when they are perpendicular ($\theta = 90^\circ$). This is the exact mirror image of the dot product, which carries $\cos\theta$ and so peaks for parallel vectors and dies for perpendicular ones.

There is a subtlety NCERT flags. Between any two vectors there are really two angles — $\theta$ and $360^\circ-\theta$. When you apply the right-hand rule you must always turn through the smaller angle, the one less than $180^\circ$. That is the $\theta$ used in $AB\sin\theta$.

The right-hand rule for direction

The magnitude $AB\sin\theta$ leaves the direction undecided — the perpendicular to a plane points either "up" or "down". The right-hand rule settles it. NCERT gives two equivalent forms. The right-handed screw rule: place a screw perpendicular to the plane of $\vec a$ and $\vec b$ and turn its head from $\vec a$ to $\vec b$ through the smaller angle; the screw advances in the direction of $\vec a\times\vec b$. The right-hand rule: open your right palm, point the fingers along $\vec a$, curl them towards $\vec b$, and your outstretched thumb points along $\vec a\times\vec b$.

Figure 1 — The right-hand rule. With $\vec A$ and $\vec B$ drawn from a common point, curling the right hand from $\vec A$ to $\vec B$ sends the thumb along $\vec A\times\vec B$, perpendicular to their plane. Reversing the order to $\vec B\times\vec A$ flips the curl and the thumb to the opposite side — the geometric statement of anti-commutativity.

The diagram makes anti-commutativity visible. Curling from $\vec B$ to $\vec A$ reverses the curl, so the thumb flips to the other side of the plane. The two products $\vec A\times\vec B$ and $\vec B\times\vec A$ have identical magnitude $AB\sin\theta$; only the sign of $\hat n$ changes. NCERT also notes a quieter property: under reflection in a mirror every ordinary (polar) vector flips sign, but $\vec a\times\vec b$ does not — it is an axial (pseudo-) vector, which is exactly the right behaviour for a quantity that names an axis of rotation.

Properties of the cross product

Four algebraic properties carry almost every NEET cross-product question. They all flow from the definition.

Property	Statement	Why it holds
Anti-commutative	$\vec A\times\vec B = -(\vec B\times\vec A)$	Same magnitude; the right-hand rule reverses the direction when the order swaps.
Distributive	$\vec A\times(\vec B+\vec C)=\vec A\times\vec B+\vec A\times\vec C$	Lets you expand a cross product component by component.
Cross with itself	$\vec A\times\vec A = \vec 0$	$\theta=0^\circ$, so $\sin 0^\circ = 0$; the magnitude is zero.
Parallel vectors	$\vec A\times\vec B = \vec 0$ if $\vec A\parallel\vec B$	$\theta=0^\circ$ or $180^\circ$, $\sin\theta=0$.
Not associative	$\vec A\times(\vec B\times\vec C)\neq(\vec A\times\vec B)\times\vec C$ in general	The bracket position changes which plane the result is perpendicular to.

Two of these deserve emphasis. Anti-commutativity is the one NEET probes most: the order of the vectors is part of the answer, not an inconvenience. $\vec A\times\vec A=\vec 0$ and the parallel-vector rule are the diagnostic that two vectors line up — if a cross product comes out zero, the vectors are parallel or anti-parallel.

Unit-vector products and the cyclic rule

The component formula is built from the cross products of the unit vectors $\hat i,\hat j,\hat k$. Two facts generate all of them. First, any unit vector crossed with itself gives zero, because the angle is $0^\circ$:

$$\hat i\times\hat i = \hat j\times\hat j = \hat k\times\hat k = \vec 0$$

Second, the three axes are mutually perpendicular, so a cross of two different unit vectors has magnitude $\sin 90^\circ = 1$ and points along the third axis. For a right-handed system the directions follow a cyclic rule:

$$\hat i\times\hat j = \hat k,\qquad \hat j\times\hat k = \hat i,\qquad \hat k\times\hat i = \hat j$$

Marching forward through the cycle $\hat i\to\hat j\to\hat k\to\hat i$ gives a positive result. Going against the cycle introduces a minus sign:

$$\hat j\times\hat i = -\hat k,\qquad \hat k\times\hat j = -\hat i,\qquad \hat i\times\hat k = -\hat j$$

Figure 2 — The cyclic mnemonic. Read in the direction of the arrows, each pair multiplies to the next unit vector with a plus sign ($\hat i\times\hat j=\hat k$); read against the arrows, the result carries a minus sign ($\hat j\times\hat i=-\hat k$). A diagonal — a unit vector crossed with itself — gives the null vector.

Component and determinant form

With the unit-vector rules in hand, the cross product of $\vec A = A_x\hat i + A_y\hat j + A_z\hat k$ and $\vec B = B_x\hat i + B_y\hat j + B_z\hat k$ follows by distributing every term. Each like-pair ($\hat i\times\hat i$) drops out, and the cross terms reassemble into the most reliable working form — a $3\times 3$ determinant with the unit vectors in the top row:

$$\vec A\times\vec B = \begin{vmatrix} \hat i & \hat j & \hat k \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} = (A_yB_z - A_zB_y)\hat i - (A_xB_z - A_zB_x)\hat j + (A_xB_y - A_yB_x)\hat k$$

The determinant is "easy to remember", as NCERT puts it, precisely because you never have to recall the cyclic signs by hand — the standard cofactor expansion supplies them. The one thing to watch is the minus sign on the middle $\hat j$ term; forgetting it is the most common arithmetic error on this topic.

Geometric meaning — area of a parallelogram

The magnitude $|\vec A\times\vec B| = AB\sin\theta$ has a clean picture. Take $\vec A$ and $\vec B$ as two adjacent sides of a parallelogram. Its base is $A$ and its height is the perpendicular drop of $\vec B$, which is $B\sin\theta$. So the area is $A\cdot B\sin\theta = |\vec A\times\vec B|$. The cross product magnitude is the area of the parallelogram spanned by the two vectors — and half of it is the area of the triangle they form.

Figure 3 — The area interpretation. With $\vec A$ and $\vec B$ as adjacent sides, the parallelogram has base $A$ and height $B\sin\theta$, so its area equals $|\vec A\times\vec B|$. The shaded triangle is exactly half: $\tfrac{1}{2}|\vec A\times\vec B|$.

This is the quickest route to an area from coordinates. Given three corner points, form two side vectors, take their cross product, and read off half its magnitude for the triangle. Worked Example 3 below does exactly this.

Dot vs cross — side by side

The two products are complementary at almost every point. The card collects the contrasts you should recall instantly in the exam hall.

Scalar product vs vector product

Dot product (·)

$\vec A\cdot\vec B = AB\cos\theta$

scalar — no direction

Result: a number (with units)
Commutative: $\vec A\cdot\vec B = \vec B\cdot\vec A$
Maximum for parallel vectors
Zero for perpendicular vectors
$\hat i\cdot\hat j = 0,\ \hat i\cdot\hat i = 1$
Physics: work $W=\vec F\cdot\vec d$, power $P=\vec F\cdot\vec v$

Cross product (×)

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

vector — perpendicular to both

Result: a vector (with an axis)
Anti-commutative: $\vec A\times\vec B = -\vec B\times\vec A$
Maximum for perpendicular vectors
Zero for parallel vectors
$\hat i\times\hat j = \hat k,\ \hat i\times\hat i = \vec 0$
Physics: torque $\vec\tau=\vec r\times\vec F$, angular momentum $\vec L=\vec r\times\vec p$

Foundations

For the full scalar-product treatment and the diagnostic of when to pick each product, see Dot Product & Cross Product in Mathematical Tools.

Why rotational quantities need it

NCERT introduces the vector product precisely here, at the gateway to rotation, because the two central quantities of the chapter cannot be built any other way. A rotation happens about an axis, and that axis is a direction that lies along neither of the vectors producing the turning effect. Only a cross product manufactures a vector perpendicular to the plane of motion, so only a cross product can carry an axial direction.

Moment of a force (torque). The turning effect of a force $\vec F$ applied at position $\vec r$ from the axis is

$$\vec\tau = \vec r\times\vec F, \qquad \tau = rF\sin\theta$$

The magnitude $rF\sin\theta$ is largest when the force is perpendicular to the arm ($\theta=90^\circ$) and zero when the force points straight along the arm ($\theta=0^\circ$) — which matches everyday experience: a push directed at the hinge of a door turns it not at all. The direction of $\vec\tau$, along the axis, encodes whether the rotation is clockwise or anticlockwise.

Angular momentum. For a particle of linear momentum $\vec p=m\vec v$ at position $\vec r$,

$$\vec L = \vec r\times\vec p, \qquad L = rp\sin\theta$$

Again the cross product supplies an axial vector — the axis about which the momentum circulates. Differentiating $\vec L=\vec r\times\vec p$ recovers $\dfrac{d\vec L}{dt}=\vec\tau$, the rotational analogue of $\vec F = d\vec p/dt$. The whole rotational dynamics of the chapter rests on this one definition.

Next step

See how these definitions drive the chapter in Torque & Angular Momentum.

Worked examples

Worked Example 1

Compute $\vec A\times\vec B$ for $\vec A = 2\hat i + 3\hat j - \hat k$ and $\vec B = \hat i - \hat j + 2\hat k$. Then find $\vec B\times\vec A$.

Set up the determinant. $$\vec A\times\vec B = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 3 & -1 \\ 1 & -1 & 2 \end{vmatrix}$$

$\hat i$ term: $(3)(2) - (-1)(-1) = 6 - 1 = 5$, so $+5\hat i$.

$\hat j$ term (subtracted): $-\big[(2)(2) - (-1)(1)\big] = -[4 + 1] = -5$, so $-5\hat j$.

$\hat k$ term: $(2)(-1) - (3)(1) = -2 - 3 = -5$, so $-5\hat k$.

Result: $\vec A\times\vec B = 5\hat i - 5\hat j - 5\hat k$. By anti-commutativity, $\vec B\times\vec A = -(\vec A\times\vec B) = -5\hat i + 5\hat j + 5\hat k$ — same magnitude $5\sqrt 3$, opposite direction.

Worked Example 2

Find a unit vector $\hat n$ perpendicular to both $\vec A = \hat i + \hat j$ and $\vec B = \hat j + \hat k$.

A vector perpendicular to both is their cross product. $$\vec A\times\vec B = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix}$$

$\hat i$: $(1)(1) - (0)(1) = 1$. $\hat j$ (subtracted): $-[(1)(1) - (0)(0)] = -1$. $\hat k$: $(1)(1) - (1)(0) = 1$. So $\vec A\times\vec B = \hat i - \hat j + \hat k$.

Normalise. Its magnitude is $\sqrt{1^2 + (-1)^2 + 1^2} = \sqrt 3$. Hence $$\hat n = \frac{\hat i - \hat j + \hat k}{\sqrt 3}.$$ The vector $-\hat n$ is equally valid — it is the perpendicular pointing to the opposite side, given by $\vec B\times\vec A$.

Worked Example 3

Find the area of the triangle with vertices $P(1,1,0)$, $Q(3,2,1)$ and $R(2,4,1)$.

Form two side vectors from $P$. $\vec{PQ} = (3-1)\hat i + (2-1)\hat j + (1-0)\hat k = 2\hat i + \hat j + \hat k$ and $\vec{PR} = \hat i + 3\hat j + \hat k$.

Cross them. $$\vec{PQ}\times\vec{PR} = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 1 & 1 \\ 1 & 3 & 1 \end{vmatrix}$$ $\hat i$: $(1)(1)-(1)(3) = -2$. $\hat j$ (subtracted): $-[(2)(1)-(1)(1)] = -1$. $\hat k$: $(2)(3)-(1)(1) = 5$. So $\vec{PQ}\times\vec{PR} = -2\hat i - \hat j + 5\hat k$.

Take half the magnitude. $|\vec{PQ}\times\vec{PR}| = \sqrt{(-2)^2 + (-1)^2 + 5^2} = \sqrt{4+1+25} = \sqrt{30}$. The triangle area is $\tfrac{1}{2}\sqrt{30} \approx 2.74$ square units. The parallelogram on the same two sides has area $\sqrt{30}$.

Quick recap

The vector product in one breath

$\vec A\times\vec B = AB\sin\theta\,\hat n$: a vector, perpendicular to both, sense by the right-hand rule.
Anti-commutative: $\vec A\times\vec B = -(\vec B\times\vec A)$. Order is part of the answer.
$\vec A\times\vec A = \vec 0$; the cross of parallel vectors is zero (uses $\sin\theta$).
Cyclic: $\hat i\times\hat j=\hat k,\ \hat j\times\hat k=\hat i,\ \hat k\times\hat i=\hat j$; reverse for a minus sign.
Compute via the $3\times3$ determinant — remember the minus sign on the $\hat j$ term.
$|\vec A\times\vec B|$ is the parallelogram area; half of it is the triangle area.
Torque $\vec\tau=\vec r\times\vec F$ and angular momentum $\vec L=\vec r\times\vec p$ need the cross product to carry an axis.

Vector Product of Two Vectors