3.16 Cross ProductHL ONLY

1. Dual Definitions of the Cross Product

The cross product operates on 3D vectors. Unlike the dot product which yields a real number, the cross product generates a new 3D vector that is perpendicular to the plane occupied by the initial vectors.

Geometric Definition:

$\vec{u} \times \vec{v} = (|\vec{u}| |\vec{v}| \sin\theta)\hat{n}$

$\theta$ is the angle between the two vectors ($0 \le \theta \le \pi$).
$\hat{n}$ is a unit vector perpendicular to both $\vec{u}$ and $\vec{v}$. Its direction follows the right-hand rule.
The cross product is anti-commutative: $\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$.

Algebraic (Determinant) Definition:

For vectors $\vec{u} = \begin{pmatrix} a_1 \\ b_1 \\ c_1 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} a_2 \\ b_2 \\ c_2 \end{pmatrix}$, the cross product is calculated using a $3 \times 3$ matrix determinant expanding along the unit vectors $\vec{i}, \vec{j}, \vec{k}$:

$\vec{u} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = \vec{i}\begin{vmatrix} b_1 & c_1 \\ b_2 & c_2 \end{vmatrix} - \vec{j}\begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} + \vec{k}\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = \begin{pmatrix} b_1 c_2 - c_1 b_2 \\ c_1 a_2 - a_1 c_2 \\ a_1 b_2 - b_1 a_2 \end{pmatrix}$

EXAMPLE 1 (Algebraic Product & Orthogonality Verification)

Compute the cross product for $\vec{u} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}$, then confirm it is perpendicular.

Step 1: Calculate the determinant:

$\vec{u} \times \vec{v} = \begin{pmatrix} (2)(6) - (3)(5) \\ (3)(4) - (1)(6) \\ (1)(5) - (2)(4) \end{pmatrix} = \begin{pmatrix} 12 - 15 \\ 12 - 6 \\ 5 - 8 \end{pmatrix} = \mathbf{\begin{pmatrix} -3 \\ 6 \\ -3 \end{pmatrix}}$

Step 2: Reversing the order of multiplication yields:

$\vec{v} \times \vec{u} = \begin{pmatrix} (5)(3) - (6)(2) \\ (6)(1) - (4)(3) \\ (4)(2) - (5)(1) \end{pmatrix} = \begin{pmatrix} 3 \\ -6 \\ 3 \end{pmatrix} = \mathbf{-\begin{pmatrix} -3 \\ 6 \\ -3 \end{pmatrix}}$

This demonstrates the anti-commutative property ($\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$).

Step 3: Verify perpendicularity using the dot product:

$(\vec{u} \times \vec{v}) \cdot \vec{u} = \begin{pmatrix} -3 \\ 6 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = -3(1) + 6(2) - 3(3) = -3 + 12 - 9 = \mathbf{0}$

Because the dot product is zero, the output vector is orthogonal.

2. Geometric Area Calculations

The magnitude of the cross product ($|\vec{u} \times \vec{v}| = |\vec{u}| |\vec{v}| \sin\theta$) equals the area of geometric figures formed by these vectors.

Area of a Parallelogram: The area of a parallelogram defined by vectors $\vec{u}$ and $\vec{v}$ is the magnitude of their cross product:

$\text{Area} = |\vec{u} \times \vec{v}|$
Area of a Triangle: The area of a triangle defined by vectors $\vec{u}$ and $\vec{v}$ is half the area of the parallelogram:

$\text{Area} = \dfrac{1}{2}|\vec{u} \times \vec{v}|$

EXAMPLE 2 (Parallelogram Area Calculation)

Calculate the area of the parallelogram defined by vectors $\vec{u} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}$.

From Example 1, we have:
$\vec{u} \times \vec{v} = \begin{pmatrix} -3 \\ 6 \\ -3 \end{pmatrix}$

The area is the magnitude of this vector:

$$\begin{aligned} \text{Area} &= \sqrt{(-3)^2 + 6^2 + (-3)^2} \\ &= \sqrt{9 + 36 + 9} \\ &= \sqrt{54} = 3\sqrt{6} \approx \mathbf{7.35\text{ units}^2} \end{aligned}$$

EXAMPLE 3 (Triangle Spanning from Spatial Points)

Calculate the area of the triangle with vertices $A(1,1,1)$, $B(1,3,1)$, and $C(-3,3,4)$.

Step 1: Find two vectors from a shared vertex (using $A$):

$\overrightarrow{AB} = \begin{pmatrix} 1-1 \\ 3-1 \\ 1-1 \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} \quad \text{and} \quad \overrightarrow{AC} = \begin{pmatrix} -3-1 \\ 3-1 \\ 4-1 \end{pmatrix} = \begin{pmatrix} -4 \\ 2 \\ 3 \end{pmatrix}$

Step 2: Calculate the cross product $\overrightarrow{AB} \times \overrightarrow{AC}$:

$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix} (2)(3) - (0)(2) \\ (0)(-4) - (0)(3) \\ (0)(2) - (2)(-4) \end{pmatrix} = \begin{pmatrix} 6 - 0 \\ 0 - 0 \\ 0 - (-8) \end{pmatrix} = \begin{pmatrix} 6 \\ 0 \\ 8 \end{pmatrix}$

Step 3: The triangle area is half the vector magnitude:

$$\begin{aligned} \text{Area} &= \dfrac{1}{2} \sqrt{6^2 + 0^2 + 8^2} \\ &= \dfrac{1}{2} \sqrt{36 + 0 + 64} \\ &= \dfrac{1}{2} \sqrt{100} = \dfrac{1}{2}(10) = \mathbf{5\text{ units}^2} \end{aligned}$$