Physics · Ch 6 — System of Particles and Rotational Motion

Vector Product of Two Vectors

6.5

Vector Product of Two Vectors

When two vectors are multiplied in a way that the result is a vector, the operation is called a vector product or cross product. Unlike scalar multiplication, which gives a number, the vector product gives a new vector whose magnitude and direction both carry physical meaning. In rotational motion, this operation appears naturally — torque, angular momentum, and angular velocity are all defined through vector products.

The vector product of two vectors $\mathbf{a}$ and $\mathbf{b}$ is written as $\mathbf{a} \times \mathbf{b}$ and read as "a cross b". The result is a third vector $\mathbf{c}$ whose magnitude is

|\mathbf{c}| = |\mathbf{a} \times \mathbf{b}| = ab \sin\theta

where $a = |\mathbf{a}|$ , $b = |\mathbf{b}|$ , and $\theta$ is the smaller angle between $\mathbf{a}$ and $\mathbf{b}$ , measured from $\mathbf{a}$ to $\mathbf{b}$ ( $0 \leq \theta \leq \pi$ ).

The direction of $\mathbf{c}$ is perpendicular to the plane containing $\mathbf{a}$ and $\mathbf{b}$ . But perpendicular to a plane gives two opposite possibilities — up or down. Which one do we choose? The answer comes from the right‑hand rule.

Tip

Right‑hand rule for direction

Stretch your right hand so that your fingers curl from $\mathbf{a}$ toward $\mathbf{b}$ through the smaller angle $\theta$ . Your extended thumb then points in the direction of $\mathbf{a} \times \mathbf{b}$ .

Because the direction depends on the order of the vectors, the vector product is not commutative. Swapping the order reverses the direction:

\mathbf{b} \times \mathbf{a} = -\,\mathbf{a} \times \mathbf{b}

This anti‑commutative property is one of the most important features of the cross product.

Watch out

Common mistake

Many students assume $\mathbf{a} \times \mathbf{b} = \mathbf{b} \times \mathbf{a}$ because scalar multiplication is commutative. That is false for vector products. The magnitudes are equal, but the directions are opposite.

Geometrical interpretation

The magnitude $ab\sin\theta$ has a simple geometric meaning. If $\mathbf{a}$ and $\mathbf{b}$ are drawn from a common origin, they form two adjacent sides of a parallelogram. The area of that parallelogram is

\text{Area} = ab\sin\theta = |\mathbf{a} \times \mathbf{b}|

So the magnitude of the cross product equals the area of the parallelogram spanned by the two vectors. If the vectors are parallel ( $\theta = 0$ or $\pi$ ), the area is zero and $\mathbf{a} \times \mathbf{b} = \mathbf{0}$ . If they are perpendicular ( $\theta = \pi/2$ ), the magnitude is simply $ab$ .

Important

Zero vector product

$\mathbf{a} \times \mathbf{b} = \mathbf{0}$ if and only if $\mathbf{a}$ and $\mathbf{b}$ are parallel (or one of them is the zero vector). This is because $\sin\theta = 0$ only when $\theta = 0$ or $\pi$ .

Vector product in terms of components

To compute cross products algebraically, we express vectors in terms of their components along the coordinate axes. Let $\mathbf{i}$ , $\mathbf{j}$ , $\mathbf{k}$ be the unit vectors along the $x$ , $y$ , $z$ axes respectively. These unit vectors are mutually perpendicular and follow the right‑hand rule: $\mathbf{i} \times \mathbf{j} = \mathbf{k}$ , $\mathbf{j} \times \mathbf{k} = \mathbf{i}$ , $\mathbf{k} \times \mathbf{i} = \mathbf{j}$ .

The cross products among the unit vectors are:

Product	Result
$\mathbf{i} \times \mathbf{i}$	$\mathbf{0}$
$\mathbf{j} \times \mathbf{j}$	$\mathbf{0}$
$\mathbf{k} \times \mathbf{k}$	$\mathbf{0}$
$\mathbf{i} \times \mathbf{j}$	$\mathbf{k}$
$\mathbf{j} \times \mathbf{k}$	$\mathbf{i}$
$\mathbf{k} \times \mathbf{i}$	$\mathbf{j}$
$\mathbf{j} \times \mathbf{i}$	$-\mathbf{k}$
$\mathbf{k} \times \mathbf{j}$	$-\mathbf{i}$
$\mathbf{i} \times \mathbf{k}$	$-\mathbf{j}$

Notice the cyclic pattern: $\mathbf{i} \to \mathbf{j} \to \mathbf{k} \to \mathbf{i}$ gives a positive result; going against the cycle gives a negative result.

Now write two vectors in component form:

\mathbf{a} = a_x \mathbf{i} + a_y \mathbf{j} + a_z \mathbf{k}, \qquad \mathbf{b} = b_x \mathbf{i} + b_y \mathbf{j} + b_z \mathbf{k}

Their vector product is

\mathbf{a} \times \mathbf{b} = (a_x \mathbf{i} + a_y \mathbf{j} + a_z \mathbf{k}) \times (b_x \mathbf{i} + b_y \mathbf{j} + b_z \mathbf{k})

Using the distributive property (which holds for vector products) and the unit‑vector results above, we expand term by term. Every product of a unit vector with itself vanishes. The cross terms give:

\begin{aligned} \mathbf{a} \times \mathbf{b} &= a_x b_y (\mathbf{i} \times \mathbf{j}) + a_x b_z (\mathbf{i} \times \mathbf{k}) \\ &\quad + a_y b_x (\mathbf{j} \times \mathbf{i}) + a_y b_z (\mathbf{j} \times \mathbf{k}) \\ &\quad + a_z b_x (\mathbf{k} \times \mathbf{i}) + a_z b_y (\mathbf{k} \times \mathbf{j}) \end{aligned}

Substituting the unit‑vector results:

\mathbf{a} \times \mathbf{b} = a_x b_y \mathbf{k} + a_x b_z (-\mathbf{j}) + a_y b_x (-\mathbf{k}) + a_y b_z \mathbf{i} + a_z b_x \mathbf{j} + a_z b_y (-\mathbf{i})

Grouping the components of $\mathbf{i}$ , $\mathbf{j}$ , $\mathbf{k}$ :

\mathbf{a} \times \mathbf{b} = (a_y b_z - a_z b_y) \mathbf{i} + (a_z b_x - a_x b_z) \mathbf{j} + (a_x b_y - a_y b_x) \mathbf{k}

This is the component form of the vector product. It is often written compactly as a determinant:

\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix}

Expanding this determinant gives exactly the same expression. The determinant form is the easiest way to remember and compute cross products.

Tip

How to use the determinant

Expand along the first row:

$\mathbf{a} \times \mathbf{b} = \mathbf{i}(a_y b_z - a_z b_y) - \mathbf{j}(a_x b_z - a_z b_x) + \mathbf{k}(a_x b_y - a_y b_x)$

The minus sign on the $\mathbf{j}$ term is crucial — it comes from the cofactor expansion.

Properties of the vector product

The textbook lists several important properties. Each one is proved directly from the definition or from the component form.

›Proof

Property 1: Anti‑commutativity

$\mathbf{a} \times \mathbf{b} = -\,\mathbf{b} \times \mathbf{a}$

From the magnitude definition, $|\mathbf{a} \times \mathbf{b}| = ab\sin\theta$ and $|\mathbf{b} \times \mathbf{a}| = ba\sin\theta$ , so the magnitudes are equal. The direction of $\mathbf{a} \times \mathbf{b}$ is given by the right‑hand rule curling from $\mathbf{a}$ to $\mathbf{b}$ ; curling from $\mathbf{b}$ to $\mathbf{a}$ reverses the thumb direction. Hence $\mathbf{b} \times \mathbf{a}$ points opposite to $\mathbf{a} \times \mathbf{b}$ , giving the negative sign.

In components, swapping the rows of the determinant changes its sign, confirming the result.

›Proof

Property 2: Distributive law

$\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}$

Write all vectors in component form. The left side becomes

$\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_x & a_y & a_z \\ b_x + c_x & b_y + c_y & b_z + c_z \end{vmatrix}$

Expanding the determinant, each term splits into a sum. For example, the $\mathbf{i}$ component is $a_y(b_z + c_z) - a_z(b_y + c_y) = (a_y b_z - a_z b_y) + (a_y c_z - a_z c_y)$ , which is exactly the sum of the $\mathbf{i}$ components of $\mathbf{a} \times \mathbf{b}$ and $\mathbf{a} \times \mathbf{c}$ . The same holds for the $\mathbf{j}$ and $\mathbf{k}$ components. Hence the distributive law holds.

›Proof

Property 3: Scalar multiplication

$(\lambda \mathbf{a}) \times \mathbf{b} = \lambda (\mathbf{a} \times \mathbf{b}) = \mathbf{a} \times (\lambda \mathbf{b})$ , where $\lambda$ is a scalar.

In the determinant, multiplying any row by $\lambda$ multiplies the entire determinant by $\lambda$ . So $(\lambda \mathbf{a}) \times \mathbf{b}$ has the first row of the determinant scaled by $\lambda$ , giving $\lambda$ times the original determinant. Similarly, scaling the second row gives $\mathbf{a} \times (\lambda \mathbf{b}) = \lambda (\mathbf{a} \times \mathbf{b})$ .

›Proof

Property 4: Parallel vectors give zero

If $\mathbf{a}$ and $\mathbf{b}$ are parallel (or one is zero), then $\mathbf{a} \times \mathbf{b} = \mathbf{0}$ .

For parallel vectors, $\theta = 0$ or $\pi$ , so $\sin\theta = 0$ and the magnitude is zero. In components, if $\mathbf{b} = \lambda \mathbf{a}$ , then the rows of the determinant are proportional, making the determinant zero.

›Proof

Property 5: Self‑product is zero

$\mathbf{a} \times \mathbf{a} = \mathbf{0}$

This is a special case of Property 4 with $\mathbf{b} = \mathbf{a}$ . The two rows of the determinant are identical, so the determinant vanishes.

›Proof

Property 6: Magnitude and dot product relation

$|\mathbf{a} \times \mathbf{b}|^2 = a^2 b^2 - (\mathbf{a} \cdot \mathbf{b})^2$

Start with $|\mathbf{a} \times \mathbf{b}|^2 = (ab\sin\theta)^2 = a^2 b^2 \sin^2\theta$ . Since $\sin^2\theta = 1 - \cos^2\theta$ , we have

$|\mathbf{a} \times \mathbf{b}|^2 = a^2 b^2 (1 - \cos^2\theta) = a^2 b^2 - (ab\cos\theta)^2$

But $ab\cos\theta = \mathbf{a} \cdot \mathbf{b}$ , so

$|\mathbf{a} \times \mathbf{b}|^2 = a^2 b^2 - (\mathbf{a} \cdot \mathbf{b})^2$

This identity connects the cross product and dot product magnitudes.

›Proof

Property 7: Scalar triple product (cyclic property)

$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) = \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b})$

…

Figure 6.15(a) Right-handed screw rule for the vector product. (b) Right-hand rule for the vector product.

Fig. 6.15 — (a) Right-handed screw rule for the vector product. (b) Right-hand rule for the vector product.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Figure 6.15 in the NCERT textbook is a pair of visual mnemonics for the vector product (also called the cross product). The figure does not plot data or show a graph; it shows two different ways to determine the direction of the vector $\mathbf{c} = \mathbf{a} \times \mathbf{b}$ when you already know the two vectors $\mathbf{a}$ and $\mathbf{b}$ and the angle $\theta$ between them.

Panel (a) uses a right-handed screw. Imagine a screw placed at the common tail of $\mathbf{a}$ and $\mathbf{b}$ , with its axis perpendicular to the plane containing $\mathbf{a}$ and $\mathbf{b}$ . If you rotate the screw from $\mathbf{a}$ toward $\mathbf{b}$ through the smaller angle $\theta$ , the direction in which the screw advances (moves forward) is the direction of $\mathbf{c} = \mathbf{a} \times \mathbf{b}$ . The figure shows the screw’s head with a curved arrow indicating the rotation, and the screw’s tip moving upward along $\mathbf{c}$ .

Panel (b) uses your right hand. Curl the fingers of your right hand from $\mathbf{a}$ toward $\mathbf{b}$ through the angle $\theta$ (the same smaller angle). Your extended thumb then points in the direction of $\mathbf{c} = \mathbf{a} \times \mathbf{b}$ . The figure shows a stylised right hand with the curled fingers and the thumb sticking out along $\mathbf{c}$ , while $\mathbf{a}$ and $\mathbf{b}$ are drawn near the wrist.

Both rules give the same result: the cross product is perpendicular to the plane containing $\mathbf{a}$ and $\mathbf{b}$ , and its sense is given by the right-hand rule. The magnitude of $\mathbf{c}$ is $|\mathbf{a}||\mathbf{b}|\sin\theta$ , which is the area of the parallelogram spanned by $\mathbf{a}$ and $\mathbf{b}$ .

\mathbf{c} = \mathbf{a} \times \mathbf{b} \quad \text{with} \quad |\mathbf{c}| = |\mathbf{a}||\mathbf{b}|\sin\theta

Watch out

The cross product is not commutative: $\mathbf{a} \times \mathbf{b} = -\,\mathbf{b} \times \mathbf{a}$ . Swapping the order reverses the direction of $\mathbf{c}$ . The right-hand rule must be applied with the first vector rotated toward the second — the screw turns from $\mathbf{a}$ to $\mathbf{b}$ , not the other way. …

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