The cross product and the right-hand rule

What you will be able to do

• state the cross product, and use the right-hand rule to find its direction;
• calculate $\vec a\times\vec b$ from components with a determinant;
• use its rules, and find a unit vector at right angles to two given vectors.

The wrench

A wrench turns a bolt. A push at right angles to the handle works best. A push along the handle does nothing. The turning effect depends on the size of the push, the length of the handle and the angle between them. It also has a direction: along the bolt, tightening or loosening. A number cannot say this, so we make a new product of two vectors that gives a vector. It is the cross product.

The definition

Let $\theta$ be the angle between two non-zero vectors $\vec a$ and $\vec b$, with $0 \le \theta \le \pi$. The cross product, also called the vector product, is

Here $\hat n$ is the unit vector at right angles to both $\vec a$ and $\vec b$, chosen by the right-hand rule. The size of the cross product is $\lvert\vec a\rvert\lvert\vec b\rvert\sin\theta$. It is zero when the vectors are parallel, because then $\sin\theta = 0$.

NCERT and IB write $\vec a\times\vec b$. A-Level books write $\mathbf a\times\mathbf b$, and some UK boards write $\mathbf a\wedge\mathbf b$. We use the cross product in three dimensions only.

The right-hand rule

Place $\vec a$ and $\vec b$ tail to tail. Curl the fingers of your right hand from $\vec a$ towards $\vec b$, through the smaller angle. Your thumb points along $\vec a\times\vec b$. Three other ways give the same direction. A right-handed screw turned from $\vec a$ to $\vec b$ moves along $\vec a\times\vec b$. The first finger, the second finger and the thumb point along $\vec a$, $\vec b$ and $\vec a\times\vec b$. For the unit vectors, $\hat i\times\hat j = \hat k$.

The rules

• $\vec b\times\vec a = -\,\vec a\times\vec b$. The order matters.
• $\vec a\times(\vec b + \vec c) = \vec a\times\vec b + \vec a\times\vec c$, and $(\lambda\vec a)\times\vec b = \lambda(\vec a\times\vec b)$.
• $\vec a\times\vec a = \vec 0$, and $\vec a\times\vec b = \vec 0$ for non-zero vectors exactly when they are parallel.
• $\hat i\times\hat j = \hat k$, $\hat j\times\hat k = \hat i$, $\hat k\times\hat i = \hat j$. Going the other way round gives a minus sign, so $\hat j\times\hat i = -\hat k$.
• The cross product is not associative: in general $(\vec a\times\vec b)\times\vec c \ne \vec a\times(\vec b\times\vec c)$.

The cross product from components

Expanding with the rules, with $\vec a = a_1\hat i + a_2\hat j + a_3\hat k$ and $\vec b = b_1\hat i + b_2\hat j + b_3\hat k$, gives the determinant

Mind the minus sign in front of the $\hat j$ term.

A unit vector at right angles to two vectors

Divide $\vec a\times\vec b$ by its size. The two unit vectors at right angles to both are $\pm\dfrac{\vec a\times\vec b}{\lvert\vec a\times\vec b\rvert}$. Use the sign that the question asks for.

In physics the torque of a force $\vec F$ about the origin is $\vec\tau = \vec r\times\vec F$, where $\vec r$ is the position of the point where the force acts. The force on a charge $q$ moving with velocity $\vec v$ in a magnetic field $\vec B$ is $q\,\vec v\times\vec B$.

A worked example

Let $\vec a = 2\hat i - \hat j + 3\hat k$ and $\vec b = \hat i + 4\hat j - 2\hat k$.

1. The $\hat i$ part. $(-1)(-2) - (3)(4) = 2 - 12 = -10$.
2. The $\hat j$ part. $-\big[(2)(-2) - (3)(1)\big] = -(-4 - 3) = 7$.
3. The $\hat k$ part. $(2)(4) - (-1)(1) = 9$. So $\vec a\times\vec b = -10\hat i + 7\hat j + 9\hat k$.
4. A check. The dot product with $\vec a$ is $-20 - 7 + 27 = 0$, and with $\vec b$ it is $-10 + 28 - 18 = 0$. So the result is at right angles to both.
5. The size. $\lvert\vec a\times\vec b\rvert = \sqrt{100 + 49 + 81} = \sqrt{230}$, so the unit vector is $\dfrac{-10\hat i + 7\hat j + 9\hat k}{\sqrt{230}}$.

Practise with new numbers

Try it in Python

The program finds $\vec a\times\vec b$ for the worked example, and checks the right angles, the change of sign and the identity of Lagrange.

Common mistakes

• Using the left hand. The thumb then points the wrong way. Check with $\hat i\times\hat j = \hat k$.
• Writing $\vec a\times\vec b = \vec b\times\vec a$. They are opposite.
• Forgetting the minus sign of the $\hat j$ term in the determinant.
• Thinking the cross product is zero only when a vector is zero. It is also zero for parallel vectors.
• Regrouping. $(\vec a\times\vec b)\times\vec c$ and $\vec a\times(\vec b\times\vec c)$ are different.

Remember these

• $\vec a\times\vec b = \lvert\vec a\rvert\lvert\vec b\rvert\sin\theta\ \hat n$, at right angles to both, by the right-hand rule.
• $\vec b\times\vec a = -\,\vec a\times\vec b$, and $\vec a\times\vec b = \vec 0$ exactly for parallel vectors.
• In components it is the determinant with rows $\hat i\ \hat j\ \hat k$, $\vec a$ and $\vec b$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What is $\hat k\times\hat i$, and what is $\hat i\times\hat k$?

2. Find $\vec a\times\vec b$ for $\vec a = \hat i + 2\hat j$ and $\vec b = 3\hat j + \hat k$.

3. Two vectors of sizes $4$ and $5$ have an angle of $30^\circ$ between them. What is the size of their cross product?

4. Find a unit vector at right angles to both $\hat i + \hat j$ and $\hat j + \hat k$.

5. If $\vec a\times\vec b = \vec a\times\vec c$ and $\vec a \ne \vec 0$, must $\vec b = \vec c$?

References

• NCERT, Mathematics Part II for Class XII, Chapter 10, section on the vector (or cross) product of vectors.
• OpenStax, Calculus Volume 3, Chapter 2, section on the cross product.