Vector (or Cross) Product of Two Vectors

Q: Is the cross product associative?

No. In general the two ways of bracketing a triple cross product give different vectors, since one lies in the plane of the first two vectors while the other lies in the plane of the last two. Brackets are compulsory in any triple product.

Q: Can I cancel a common vector across a cross product equation?

Not in the ordinary sense. From a cross b equals a cross c you may only conclude that a cross the difference of b and c is zero, which means that difference is some scalar multiple of a.

MathsVector AlgebraFor JEE aspirants

Vector (or Cross) Product of Two Vectors

Maths · Vector Algebra · Concept 3

The cross product takes two vectors and returns a third vector, perpendicular to both, whose length measures the area they sweep out. It is the tool for areas, for normals to planes, for testing parallelism, and in physics for torque and angular velocity. Unlike the dot product it is not commutative, and that sign change is where most errors happen.

Definition

Vector product

For non-zero vectors $a$ and $b$ with angle $θ$ between them, where $0 \leq θ \leq π$ ,

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

Here $\overset{n}{^}$ is the unit vector perpendicular to the plane of $a$ and $b$ , directed so that $a$ , $b$ , $\overset{n}{^}$ form a right handed system.

If either vector is $0$ , or if the two are parallel, then $a \times b = 0$ .

Right hand rule. Curl the fingers of your right hand from $a$ towards $b$ through the angle $θ$ ; the extended thumb points along $\overset{n}{^}$ . Reversing the order reverses the thumb, which is why $b \times a = - (a \times b)$ .

The magnitude of the cross product equals the area of the parallelogram with the two vectors as adjacent sides, and its direction is the normal given by the right hand rule.

Geometrical meaning: area

Taking $a$ as the base, the height of the parallelogram is $∣ b ∣ sin θ$ , so the area is $∣ a ∣ ∣ b ∣ sin θ = ∣ a \times b ∣$ .

Area formulas

Parallelogram with adjacent sides $a$ and $b$ : $Area = ∣ a \times b ∣$

Triangle with two sides $a$ and $b$ from one vertex: $Area = \frac{1}{2} ∣ a \times b ∣$

Triangle with vertices $A (a), B (b), C (c)$ : $Area = \frac{1}{2} a \times b + b \times c + c \times a$

Quadrilateral with diagonals $d_{1}$ and $d_{2}$ : $Area = \frac{1}{2} d_{1} \times d_{2}$

Collinearity from the area

Three points $A, B, C$ are collinear exactly when the triangle they form has zero area, that is

$a \times b + b \times c + c \times a = 0$ .

Products of the base vectors and the determinant form

Base vector products

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

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

Reversing any of these reverses the sign: $\hat{j} \times \hat{i} = - \hat{k}$ , and so on. The cyclic order $i \to j \to k \to i$ gives a plus sign, the anticyclic order gives a minus sign.

Determinant form

If $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ and $b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ , then

a \times b = \hat{i} \hat{j} \hat{k} a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} = (a_{2} b_{3} - a_{3} b_{2}) \hat{i} - (a_{1} b_{3} - a_{3} b_{1}) \hat{j} + (a_{1} b_{2} - a_{2} b_{1}) \hat{k}

Properties

Algebraic properties

$a \times b = - (b \times a)$ (anticommutative, so the order is never free)
$a \times (b + c) = a \times b + a \times c$ (distributive)
$(m a) \times b = a \times (m b) = m (a \times b)$ for a scalar $m$
$a \times a = 0$
$a \times b = 0$ with $a \neq = 0, b \neq = 0$ $⟺$ $a ∥ b$ , that is $a = k b$
The cross product is not associative: in general $(a \times b) \times c \neq = a \times (b \times c)$

Normals and angles

Unit vector perpendicular to the plane of $a$ and $b$ : $\overset{n}{^} = \pm \frac{a \times b}{a \times b}$

Vector of magnitude $r$ perpendicular to that plane: $\pm \frac{r ( a \times b )}{a \times b}$

Sine of the angle: $sin θ = \frac{a \times b}{∣ a ∣ ∣ b ∣}$

Swapping the order of the two vectors flips the normal to the opposite side of the plane.

Example 1

Find a vector of magnitude $9$ perpendicular to both $4 \hat{i} - \hat{j} + 3 \hat{k}$ and $- 2 \hat{i} + \hat{j} - 2 \hat{k}$ .

Solution.

a \times b = \hat{i} \hat{j} \hat{k} 4 - 13 - 21 - 2 = (2 - 3) \hat{i} - (- 8 + 6) \hat{j} + (4 - 2) \hat{k} = - \hat{i} + 2 \hat{j} + 2 \hat{k}

$a \times b = 1 + 4 + 4 = 3$ , so the required vectors are

$\pm 9 \cdot \frac{- i ^ + 2 j ^ + 2 k ^}{3} = \pm (- 3 \hat{i} + 6 \hat{j} + 6 \hat{k})$ .

Example 2

Prove that $a \times (b + c) + b \times (c + a) + c \times (a + b) = 0$ .

Solution. Expanding by the distributive law,

$a \times b + a \times c + b \times c + b \times a + c \times a + c \times b$ .

The terms pair off, since $b \times a = - a \times b$ , $c \times a = - a \times c$ and $c \times b = - b \times c$ . The sum is $0$ .

Example 3

For any vector $a$ , prove that $a \times \hat{i}^{2} + a \times \hat{j}^{2} + a \times \hat{k}^{2} = 2∣ a ∣^{2}$ .

Solution. Let $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ . Then

$a \times \hat{i} = a_{2} (\hat{j} \times \hat{i}) + a_{3} (\hat{k} \times \hat{i}) = - a_{2} \hat{k} + a_{3} \hat{j}$ , so $a \times \hat{i}^{2} = a_{2}^{2} + a_{3}^{2}$ .

Similarly $a \times \hat{j}^{2} = a_{1}^{2} + a_{3}^{2}$ and $a \times \hat{k}^{2} = a_{1}^{2} + a_{2}^{2}$ .

Adding gives $2 (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) = 2∣ a ∣^{2}$ .

Lagrange's identity

a \times b^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2} = a \cdot a a \cdot b a \cdot b b \cdot b

Equivalently $a \times b^{2} + (a \cdot b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2}$ , which is just $sin^{2} θ + cos^{2} θ = 1$ in disguise.

How to use it

Whenever a question gives you $∣ a ∣$ , $∣ b ∣$ and one of $a \cdot b$ or $a \times b$ , this identity hands you the other immediately, with no need to find $θ$ .

Example 4

$p$ and $q$ are unit vectors with an angle of $3 0^{\circ}$ between them. Find the area of the parallelogram whose diagonals are $a = p + 2 q$ and $b = 2 p + q$ .

Solution. Area with given diagonals is $\frac{1}{2} a \times b$ .

$a \times b = (p + 2 q) \times (2 p + q) = p \times q + 4 (q \times p) = - 3 (p \times q)$ .

$∣ p \times q ∣ = 1 \cdot 1 \cdot sin 3 0^{\circ} = \frac{1}{2}$ , so $a \times b = \frac{3}{2}$ .

Area $= \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4}$ square units.

Example 5

$O A = a$ , $O B = 10 a + 2 b$ , $O C = b$ , with $O$ the origin. If $p$ is the area of quadrilateral $O A B C$ and $q$ is the area of the parallelogram with $O A$ and $O C$ as adjacent sides, prove $p = 6 q$ .

Solution. The diagonals of $O A B C$ are $O B$ and $A C = b - a$ , so

$p = \frac{1}{2} (10 a + 2 b) \times (b - a) = \frac{1}{2} 10 (a \times b) - 2 (b \times a) = \frac{1}{2} 12 (a \times b) = 6 a \times b$ .

Also $q = a \times b$ . Hence $p = 6 q$ .

JEE Advanced

Normal to the plane through three points. If $A, B, C$ have position vectors $a, b, c$ , then two sides of the triangle are $b - a$ and $c - a$ , and

$(b - a) \times (c - a) = b \times c - b \times a - a \times c = a \times b + b \times c + c \times a$ .

So the normal to the plane $A B C$ lies along $a \times b + b \times c + c \times a$ , and the area of the triangle is half its magnitude. This single expression is symmetric in the three points, which makes it far more convenient than picking a vertex.

JEE Advanced

Solving a vector equation. To solve $r \times b = c \times b$ for $r$ , rewrite it as $(r - c) \times b = 0$ , which says $r - c$ is parallel to $b$ . Hence

$r = c + k b$ , $k \in R$ .

A cross product equation alone never has a unique solution, because a whole line of vectors satisfies it. A second condition such as $r \cdot a = 0$ is needed to pin down $k$ , and substituting gives $k = - \frac{a \cdot c}{a \cdot b}$ , valid when $a$ is not perpendicular to $b$ .

Where the cross product is used

Moment (torque) of a force $F$ acting at a point with position vector $r$ about the origin is $M = r \times F$ .
Velocity in circular motion is $v = ω \times r$ , where $ω$ is the angular velocity.
Normals to planes, and hence the equation of a plane through three points.
Areas of triangles, parallelograms and quadrilaterals from their vertices or diagonals.
Shortest distance between skew lines, whose direction is along the cross product of the two direction vectors.

Mistakes that cost marks

Writing $a \times b = b \times a$ . The sign flips, and this single slip reverses normals, areas orientations and torques.
Expanding brackets as if the product were associative. Keep the order of every factor exactly as it stands.
Setting $∣ a \times b ∣ = ∣ a ∣∣ b ∣ cos θ$ . It is $sin θ$ here, and $cos θ$ for the dot product.
Concluding $b = c$ from $a \times b = a \times c$ . All that follows is $b - c$ is parallel to $a$ .
Forgetting the $\pm$ when a question asks for "a vector perpendicular to both". There are always two opposite answers unless the orientation is specified.
Dropping the middle sign in the determinant expansion. The $\hat{j}$ term carries a minus sign.

Quick recap

$a \times b = ∣ a ∣∣ b ∣ sin θ \overset{n}{^}$ , with $\overset{n}{^}$ from the right hand rule
Determinant form with rows $\hat{i}, \hat{j}, \hat{k}$ ; then $a$ ; then $b$
$a ∥ b ⟺ a \times b = 0$ ; $a \times a = 0$ ; $b \times a = - a \times b$
Areas: parallelogram $∣ a \times b ∣$ , triangle $\frac{1}{2} ∣ a \times b ∣$ , quadrilateral $\frac{1}{2} ∣ d_{1} \times d_{2} ∣$
Lagrange: $a \times b^{2} + (a \cdot b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2}$
$M = r \times F$ , $v = ω \times r$

Frequently asked questions

Why is the cross product defined only in three dimensions?

Because the defining requirement is a vector perpendicular to two given vectors. In three dimensions that direction is unique up to sign, in two dimensions no such vector lies in the plane, and in higher dimensions the perpendicular directions form a space of more than one dimension, so no single answer exists.

Is the cross product associative?

No. In general $(a \times b) \times c$ and $a \times (b \times c)$ are different vectors, since the first lies in the plane of $a$ and $b$ while the second lies in the plane of $b$ and $c$ . Brackets are compulsory in any triple product.

How do I choose between the two possible normal directions?

The right hand rule fixes it once the order of the two vectors is fixed: curl your fingers from the first vector to the second and your thumb gives the normal. If the question just asks for a perpendicular vector without specifying orientation, both signs are acceptable answers.

Can I cancel a common vector across a cross product equation?

Not in the ordinary sense. From $a \times b = a \times c$ you may only conclude $a \times (b - c) = 0$ , which means $b - c$ is some scalar multiple of $a$ , so $b = c + k a$ .

When should I use the cross product rather than the dot product to find an angle?

Use the dot product as the default, because the cosine determines the angle uniquely on the range from zero to $π$ . The sine from the cross product cannot distinguish $θ$ from $π - θ$ , so it is safe only when you already know whether the angle is acute or obtuse.

Previous year questions on Vector (or Cross) Product of Two Vectors

17 questions from past papers, each with a step-by-step solution.

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