Cross product Knowledge-Based Study

Study notes

1. What is the Cross Product?

   In physics, certain operations require combining two vectors to produce a third vector that is perpendicular to both original vectors. This vector operation is known as the vector product or cross product.

   If two vectors $\vec{A}$ and $\vec{B}$ are placed tail-to-tail with an angle $\theta$ between them (where $0^\circ \le \theta \le 180^\circ$), their cross product is written as:

   $$\vec{C} = \vec{A} \times \vec{B}$$

   Unlike the scalar (dot) product, the result of a cross product is always a vector quantity.

   Magnitude of the Cross Product

   The magnitude of $\vec{C}$ is given by:

   $$|\vec{C}| = |\vec{A} \times \vec{B}| = A B \sin\theta$$

   where $A = |\vec{A}|$ and $B = |\vec{B}|$.

   • Parallel or Antiparallel Vectors: If $\vec{A}$ and $\vec{B}$ are parallel ($\theta = 0^\circ$) or antiparallel ($\theta = 180^\circ$), $\sin\theta = 0$, so $|\vec{A} \times \vec{B}| = 0$. In particular, the cross product of any vector with itself is zero: $$\vec{A} \times \vec{A} = 0$$
   • Perpendicular Vectors: If $\vec{A}$ and $\vec{B}$ are mutually perpendicular ($\theta = 90^\circ$), $\sin 90^\circ = 1$, which maximizes the magnitude: $$|\vec{A} \times \vec{B}| = A B$$

   Geometric Interpretation

   The magnitude $|\vec{A} \times \vec{B}|$ equals the area of the parallelogram formed by the vectors $\vec{A}$ and $\vec{B}$. Consequently, the area of the triangle spanned by $\vec{A}$ and $\vec{B}$ is equal to $\frac{1}{2}|\vec{A} \times \vec{B}|$.

   Determining Direction

   The vector resulting from a cross product, $\vec{C} = \vec{A} \times \vec{B}$, is directed perpendicular to the plane formed by $\vec{A}$ and $\vec{B}$. Because there are two opposite directions perpendicular to any plane, we determine the unique direction using the right-hand rule.

   Applying the Right-Hand Rule

   1. Extend the fingers of your right hand in the direction of the first vector, $\vec{A}$.
   2. Curl your fingers toward the direction of the second vector, $\vec{B}$, through the smallest angle $\theta$ between them.
   3. Your extended thumb points in the direction of the cross product vector, $\vec{C}$.

   Key Algebraic Properties

   • Anti-commutativity: Reversing the order of vectors reverses the direction of the resulting vector: $$\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})$$
   • Distributive Property: The cross product distributes over vector addition: $$\vec{A} \times (\vec{B} + \vec{C}) = (\vec{A} \times \vec{B}) + (\vec{A} \times \vec{C})$$
   • Scalar Multiplication: For any scalar $k$: $$(k\vec{A}) \times \vec{B} = \vec{A} \times (k\vec{B}) = k(\vec{A} \times \vec{B})$$
   • Non-associativity: The cross product is generally not associative: $$\vec{A} \times (\vec{B} \times \vec{C}) \neq (\vec{A} \times \vec{B}) \times \vec{C}$$

   Unit Vectors in Cartesian Coordinates

   Consider the standard Cartesian unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$ pointing along the positive $x$-, $y$-, and $z$-axes respectively in a right-handed coordinate system.

   Using the definition of the cross product:

   • $\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = 0$
   • $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$
   • $\hat{j} \times \hat{i} = -\hat{k}$, $\hat{k} \times \hat{j} = -\hat{i}$, $\hat{i} \times \hat{k} = -\hat{j}$

   A helpful cyclic mnemonic is the order $\hat{i} \rightarrow \hat{j} \rightarrow \hat{k} \rightarrow \hat{i}$. Going forward along the cycle gives a positive sign; going backward gives a negative sign.

   Component Form via Determinants

   When vectors are expressed in component form, $\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$ and $\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}$, their cross product can be computed using a $3 \times 3$ matrix determinant:

   $$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}$$

   Expanding across the first row yields the individual components:

   $$\vec{A} \times \vec{B} = (A_y B_z - A_z B_y)\hat{i} - (A_x B_z - A_z B_x)\hat{j} + (A_x B_y - A_y B_x)\hat{k}$$

   Notice the negative sign associated with the $\hat{j}$ cofactor, which can also be written as $+(A_z B_x - A_x B_z)\hat{j}$.

   Physical Applications

   The cross product appears across classical mechanics and electromagnetism whenever a physical quantity depends on perpendicular spatial relationships and rotational orientations:

   • Torque ($\vec{\tau}$): The rotational analog of force is given by $\vec{\tau} = \vec{r} \times \vec{F}$, where $\vec{r}$ is the position vector from the pivot point to the line of action of force $\vec{F}$.
   • Angular Momentum ($\vec{L}$): For a particle, $\vec{L} = \vec{r} \times \vec{p}$, where $\vec{p}$ is linear momentum.
   • Magnetic Force ($\vec{F}_B$): The force on a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is $\vec{F}_B = q(\vec{v} \times \vec{B})$.

   Worked Example

   Problem: Given $\vec{r} = (2\hat{i} + 3\hat{j})\text{ m}$ and $\vec{F} = (4\hat{j} + 5\hat{k})\text{ N}$, calculate the torque $\vec{\tau} = \vec{r} \times \vec{F}$.

   Solution:

   1. Identify the components:
      $r_x = 2$, $r_y = 3$, $r_z = 0$
      $F_x = 0$, $F_y = 4$, $F_z = 5$
   2. Set up the determinant expansion:
      $\tau_x = r_y F_z - r_z F_y = (3)(5) - (0)(4) = 15\text{ N}\cdot\text{m}$
      $\tau_y = -(r_x F_z - r_z F_x) = -[(2)(5) - (0)(0)] = -10\text{ N}\cdot\text{m}$
      $\tau_z = r_x F_y - r_y F_x = (2)(4) - (3)(0) = 8\text{ N}\cdot\text{m}$
   3. Write the final vector:
      $$\vec{\tau} = (15\hat{i} - 10\hat{j} + 8\hat{k})\text{ N}\cdot\text{m}$$