A mechanic in a Pune garage loosens the wheel nut of a car with a spanner \(25\,\text{cm}\) long. A push of \(40\,\text{N}\) at the end, at right angles to the spanner, turns the nut. The same push at a slant turns it less, and a push along the spanner does nothing. Pushing the end up loosens the nut; pushing it down tightens it. The turning effect has a size and a sense, and it depends on two vectors at once.
A product that is a vector
The second product of two vectors gives a vector. It is called the cross product, or the vector product, and it is written with a cross. Its magnitude is the product of the two magnitudes and the sine of the angle between them. So it is largest when the two vectors are perpendicular and zero when they are parallel. That is exactly the pattern of the spanner.
The cross product of two vectors \(\vec A\) and \(\vec B\) at an angle \(\theta\) is the vector \(\vec A\times\vec B\) of magnitude \(AB\sin\theta\). Its direction is at right angles to both \(\vec A\) and \(\vec B\), given by the right-hand rule. When the fingers of the right hand turn from \(\vec A\) towards \(\vec B\) through the smaller angle, the thumb points along \(\vec A\times\vec B\).
An area with a direction
The magnitude \(AB\sin\theta\) is an area. In the parallelogram with sides \(\vec A\) and \(\vec B\), take \(\vec A\) as the base. The height is \(B\sin\theta\). So the area is \(AB\sin\theta\), the magnitude of the cross product. The direction of the cross product is the normal to that parallelogram. So a cross product is an area with a direction.
The right-hand rule and the order
The right-hand rule fixes the direction. Point the fingers of your right hand along \(\vec A\), and curl them towards \(\vec B\) through the smaller angle. The stretched thumb points along \(\vec A\times\vec B\). Swap the order, and the fingers curl the other way. So the thumb reverses, and \(\vec B\times\vec A = -\vec A\times\vec B\). Unlike the dot product, the cross product depends on the order.
In a problem drawn on the page, both vectors lie in the page and their cross product sticks out of it. A dot in a circle, \(\odot\), marks a vector pointing out of the page, towards you. A cross in a circle, \(\otimes\), marks a vector pointing into the page. Think of an arrow seen from its tip and from its tail feathers.
A spanner lies along the page from a nut to the right. A force at its end pushes it downward, in the page, so it turns clockwise. Where does the torque \(\vec r\times\vec F\) point?
A force acts along the line of the spanner itself, pulling the end straight away from the nut. What is the torque \(\vec r\times\vec F\)?
The unit vectors in a cycle
The axes \(x\), \(y\) and \(z\) are chosen so that the right-hand rule links them. Curl the fingers from \(\hat i\) towards \(\hat j\), and the thumb points along \(\hat k\). Each of the three unit vectors has magnitude 1, and they meet at \(90°\), so each cross product has magnitude \(1 \times 1 \times \sin 90° = 1\). Going round the cycle \(x\) to \(y\) to \(z\) and back to \(x\) gives plus signs.
Against the cycle, each product changes sign: \(\hat j\times\hat i = -\hat k\), \(\hat k\times\hat j = -\hat i\) and \(\hat i\times\hat k = -\hat j\). Any vector crossed with itself gives the zero vector, because the angle is zero. When the \(x\) axis points right and the \(y\) axis up the page, \(\hat k\) points out of the page.
What is \(\hat i\times\hat k\)?
The cross product from components
The cycle turns the cross product into arithmetic. Write both vectors in components and multiply out the brackets. The three products of a unit vector with itself vanish. The other six products give \(\hat i\), \(\hat j\) or \(\hat k\), with a sign from the cycle. The derivation below collects them. In a plane, where \(A_z = B_z = 0\), only the \(\hat k\) term survives.
\[ \vec A\times\vec B = (A_x\hat i + A_y\hat j + A_z\hat k)\times(B_x\hat i + B_y\hat j + B_z\hat k) \]
Why. write both vectors in components. The brackets may be multiplied out, provided the order of each pair is kept.
\[ \hat i\times\hat i = \hat j\times\hat j = \hat k\times\hat k = \vec 0 \]
Why. a vector crossed with itself is zero, so three of the nine products vanish.
\[ A_xB_y\,\hat k - A_yB_x\,\hat k \]
Why. \(\hat i\times\hat j = \hat k\) and \(\hat j\times\hat i = -\hat k\) collect into the \(\hat k\) term.
\[ A_yB_z\,\hat i - A_zB_y\,\hat i, \quad A_zB_x\,\hat j - A_xB_z\,\hat j \]
Why. \(\hat j\times\hat k = \hat i\), \(\hat k\times\hat j = -\hat i\), \(\hat k\times\hat i = \hat j\) and \(\hat i\times\hat k = -\hat j\) give the other two terms.
\[ \vec A\times\vec B = (A_yB_z - A_zB_y)\hat i + (A_zB_x - A_xB_z)\hat j + (A_xB_y - A_yB_x)\hat k \]
Read it. add the three terms. Each uses the two components that are not its own, in the cyclic order.
A determinant is only a way to remember the pattern. Expand it along the top row. For the \(\hat i\) term, cover the row and the column of \(\hat i\) and cross-multiply what is left: \(A_yB_z - A_zB_y\). The \(\hat j\) term takes a minus sign in front, and the \(\hat k\) term a plus. For vectors in a plane the whole product is \((A_xB_y - A_yB_x)\hat k\). For example, \(\vec A = 4\hat i\) and \(\vec B = 3\hat i + 2\hat j\) give \(\vec A\times\vec B = 8\hat k\), a parallelogram of area 8.
This is the worked example "A torque in three dimensions" of Torque and Equilibrium, lesson 2, done with the tool. A force \(\vec F = (2\hat i - \hat j + \hat k)\,\text{N}\) acts at the point \(\vec r = (\hat i + 2\hat j + 3\hat k)\,\text{m}\). Find its torque \(\vec\tau = \vec r\times\vec F\) about the origin and the size of that torque.
\(x = 1\), \(y = 2\), \(z = 3\) in metres, and \(F_x = 2\), \(F_y = -1\), \(F_z = 1\) in newtons. The first vector is \(\vec r\), so its components form the middle row of the determinant.
\(\tau_x = yF_z - zF_y\), \(\tau_y = zF_x - xF_z\), \(\tau_z = xF_y - yF_x\).
\(\tau_x = 2 \times 1 - 3 \times (-1) = 5\). \(\tau_y = 3 \times 2 - 1 \times 1 = 5\). \(\tau_z = 1 \times (-1) - 2 \times 2 = -5\).
\(\vec\tau = (5\hat i + 5\hat j - 5\hat k)\,\text{N m}\), of size \(\sqrt{75} = 8.66\,\text{N m}\), the same answer as the physics chapter. Check with a dot product: \(\vec\tau\cdot\vec F = 10 - 5 - 5 = 0\), so the torque is perpendicular to the force, as a cross product must be.
A charge of \(2\,\mu\text{C}\) moves with \(\vec v = (3\hat i + 4\hat j)\times10^4\,\text{m/s}\) in a magnetic field \(\vec B = 0.1\,\hat k\,\text{T}\). The force is \(\vec F = q\,\vec v\times\vec B\). Find the size of the force, in mN.
A ball is at the point \((4\,\text{m}, 3\,\text{m})\) with momentum \(\vec p = (1\hat i + 2\hat j)\,\text{kg m/s}\). Its angular momentum about the origin is \(\vec L = \vec r\times\vec p\). Find \(L_z\), in kg m²/s.
\(AB\sin\theta\) gives only the size. Since \(\theta\) lies between \(0°\) and \(180°\), the sine is never negative and carries no sign. The direction must come from the right-hand rule or from the components. The order always matters: \(\vec r\times\vec F\) is the torque, and \(\vec F\times\vec r\) is its negative. A cross product needs three dimensions. For two vectors in the page it points out of the page or into it, never along the page.
Three vectors can be combined too. The scalar triple product \(\vec A\cdot(\vec B\times\vec C)\) is a number, the volume of the box with edges \(\vec A\), \(\vec B\) and \(\vec C\). It is zero exactly when the three vectors lie in one plane. The vector triple product obeys \(\vec A\times(\vec B\times\vec C) = \vec B(\vec A\cdot\vec C) - \vec C(\vec A\cdot\vec B)\). The physics of this book does not need either of them. They appear in entrance examinations and in engineering mechanics.
William Rowan Hamilton was walking with his wife along the Royal Canal in Dublin. For years he had tried to multiply triples of numbers, and he had failed. On the walk he saw that he needed four numbers, and a rule that he later called quaternions. He cut the rule \(i^2 = j^2 = k^2 = ijk = -1\) into the stone of Broome Bridge with his knife. From it follow \(ij = k\), \(jk = i\) and \(ki = j\), with a minus sign in the other order. The cross product of this lesson keeps that cycle.
The wheel nut of the opening can be done. The torque is \(\vec\tau = \vec r\times\vec F\), with \(r = 0.25\,\text{m}\) and \(F = 40\,\text{N}\). At right angles its size is \(0.25 \times 40 \times \sin 90° = 10\,\text{N m}\). At \(30°\) to the spanner it is half that, \(5\,\text{N m}\), and along the spanner it is zero. Up or down decides whether the torque points out of the wheel or into it, which loosens or tightens the nut.
Torque and Equilibrium defines torque as \(\vec r\times\vec F\), and Angular Momentum defines \(\vec L = \vec r\times\vec p\), whose rate of change is the torque. Kinematics of Circular Motion writes the velocity on a circle as \(\vec v = \vec\omega\times\vec r\), and Rolling Motion adds such a velocity to the velocity of the centre. Magnetic Force on Moving Charges gives the force \(q\,\vec v\times\vec B\) and finds its direction by the right-hand rule. The force on a wire, the torque on a loop, the Biot-Savart law and the emf of a moving rod are all cross products.
- Reversing the order of a cross product.\(\vec B\times\vec A = -\vec A\times\vec B\). Keep the order of the physics: \(\vec r\) before \(\vec F\), \(\vec v\) before \(\vec B\).
- Using the left hand, or curling the fingers from the second vector to the first.Right hand only, fingers from the first vector towards the second through the smaller angle.
- Writing \(\hat i\times\hat k = \hat j\).The cycle runs \(\hat k\times\hat i = \hat j\). Against the cycle the sign changes, so \(\hat i\times\hat k = -\hat j\).