Understanding the cross product

Cross product is a mathematical tool that helps us encapsulate the following properties:

  • The resulting vector’s direction is perpendicular to both original vectors.
  • The resulting vector’s magnitude is proportional to the area of the parallelogram formed by the two vectors.

Key properties of the cross product include:

  • Anticommutativity: The order of multiplication matters and reverses the direction of the resulting vector. \(\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})\)

  • Distributivity over addition: \((\vec{a} + \vec{b}) \times \vec{c} = (\vec{a} \times \vec{c}) + (\vec{b} \times \vec{c})\)

  • Vector triple product: \(\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b})\)

Component-wise Calculation

Given two vectors: \(\vec{a} = \begin{bmatrix} a_x \\ a_y \\ a_z \end{bmatrix}, \quad \vec{b} = \begin{bmatrix} b_x \\ b_y \\ b_z \end{bmatrix}\)

The cross product, \(\vec{c} = \vec{a} \times \vec{b}\), is calculated as: \(\vec{c} = \begin{bmatrix} a_y b_z - a_z b_y \\ a_z b_x - a_x b_z \\ a_x b_y - a_y b_x \end{bmatrix}\)

Geometric Interpretation: Magnitude and Area

The magnitude of the cross product is defined as: \(|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(\theta)\) where \(\theta\) is the angle between vectors \(\vec{a}\) and \(\vec{b}\).

This magnitude has a direct geometric meaning: it is the area of the parallelogram formed by the two vectors.

We can understand this from the formula for the area of a parallelogram: \(\text{Area} = \text{base} \times \text{height}\)

If we consider vector \(\vec{a}\) as the base, its length is \(\|\vec{a}\|\). The height of the parallelogram is the component of \(\vec{b}\) perpendicular to \(\vec{a}\), which is \(\|\vec{b}\| \sin(\theta)\).

Therefore, the area is: \(\text{Area} = |\vec{a}| (|\vec{b}| \sin(\theta)) = |\vec{a}| |\vec{b}| \sin(\theta)\) This confirms that the magnitude of the cross product equals the area of the parallelogram.

Application

Now that we understand the cross product, what are its applications? Let’s explore one in the context of rigid body rotation.

Physical Intuition

To build our intuition, let’s start with a simple physical example. If we tie a stone to a thread and spin it, we observe circular motion.

  • Axis of Rotation: Our fingers holding the thread act as the axis of rotation.
  • Radius: The thread’s length, \(r\), is the radius of the circular path.
  • Point Mass: The stone, \(q\), can be treated as a point mass.

Let’s model this mathematically by considering an instantaneous moment. Assume the axis of rotation is along the z-axis, and at a particular instant, the string is aligned with the x-axis.

The stone is moving, which means it has a velocity. But in which direction is this velocity vector pointing?

  1. It’s not along the x-axis (the direction of the string), because if it were, the string’s length would change.
  2. It’s not along the z-axis (the axis of rotation), as there is no vertical movement.

If we let go of the thread, the stone flies off in a straight line, tangent to its circular path at that instant. This tells us the instantaneous velocity is tangential to the radius.

In uniform circular motion, an inward (centripetal) force, provided by the tension in the thread, is always perpendicular to the tangential velocity, which keeps the object moving in a circle.

The Role of the Cross Product

From this, we can conclude that the instantaneous velocity vector is perpendicular to both:

  1. The rotation axis vector (e.g., along the z-axis).
  2. The radius vector (from the axis to the stone, e.g., along the x-axis).

This is where the cross product becomes incredibly useful. It gives us exactly what we need: a new vector that is perpendicular to two original vectors. The relationship between angular velocity (\(\vec{\omega}\)), the position vector of the point (\(\vec{r}\)), and its linear velocity (\(\vec{v}\)) is elegantly captured by the cross product:

\[\vec{v} = \vec{\omega} \times \vec{r}\]

This equation models the tangential velocity of a point on a rotating rigid body.

Let’s consider two cases for the point’s location to see how this formula works:

  1. The point is at a distance from the axis of rotation (like the stone at the end of the rope). Here, the position vector \(\vec{r}\) is not parallel to the angular velocity vector \(\vec{\omega}\). The cross product \(\vec{\omega} \times \vec{r}\) yields a non-zero velocity vector \(\vec{v}\), representing the linear motion of the point.

  2. The point is on the axis of rotation. In this case, the position vector \(\vec{r}\) is either the zero vector or is parallel to the angular velocity vector \(\vec{\omega}\). The cross product of two parallel vectors is the zero vector (since \(\|\vec{\omega}\|, \|\vec{r}\| \sin(0) = 0\)). Therefore, \(\vec{v} = \vec{\omega} \times \vec{r} = \vec{0}\). This correctly models that points on the axis of rotation have no linear velocity; they only rotate in place.