3.16 Vector (or Cross) Product (HL)

1. Dual Definitions of the Cross Product

The cross product algorithm operates strictly on 3D vectors. Unlike scalar multiplication, it generates an entirely new 3D spatial vector inherently perpendicular to the structural plane occupied by the initial operands.

Geometric Definition:

$\vec{u} \times \vec{v} = (|\vec{u}| |\vec{v}| \sin\theta)\hat{n}$
  • $\theta$ is the included bounded angle ($0 \le \theta \le \pi$).
  • $\hat{n}$ acts as a spatial unit vector resting perpendicularly to both $\vec{u}$ and $\vec{v}$. Directionality adheres strictly to the "right-hand rule" (or screw rule).
  • Crucially, operations are anti-commutative: $\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$.

Algebraic (Determinant) Definition:

Given vectors $\vec{u} = \binom{a_1}{b_1}{c_1}$ and $\vec{v} = \binom{a_2}{b_2}{c_2}$, the product matrix evaluates utilizing determinants:

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

EXAMPLE 1

Compute evaluating structures for $\vec{u} = \binom{1}{2}{3}$ and $\vec{v} = \binom{4}{5}{6}$.

Applying the algebraic determinant algorithm:
$\vec{u} \times \vec{v} = \binom{(2)(6) - (3)(5)}{(3)(4) - (1)(6)}{(1)(5) - (2)(4)} = \binom{12 - 15}{12 - 6}{5 - 8} = \mathbf{\binom{-3}{6}{-3}}$.
Reversing the multiplier variables equates to $\vec{v} \times \vec{u} = \mathbf{\binom{3}{-6}{3}}$, validating the anti-commutative principle mathematically.
Verifying perpendicular states utilizing the scalar dot product:
$(\vec{u} \times \vec{v}) \cdot \vec{u} = -3(1) + 6(2) - 3(3) = -3 + 12 - 9 = \mathbf{0}$. The output geometric vector rests perpendicularly intact.

2. Geometric Area Calculations

Analyzing the geometric cross product definition ($|\vec{u} \times \vec{v}| = |\vec{u}| |\vec{v}| \sin\theta$) mirrors precisely the area calculation for polygons spanning the targeted vectors.

  • Area of a Parallelogram: Supported geometrically by vectors $\vec{u}$ and $\vec{v}$ evaluates to $\mathbf{|\vec{u} \times \vec{v}|}$.
  • Area of a Triangle: Bounded between vectors $\vec{u}$ and $\vec{v}$ evaluates to strictly half the parallelogram dimension: $\mathbf{\frac{1}{2}|\vec{u} \times \vec{v}|}$.

EXAMPLE 3

Extract the physical operational area for the geometrical parallelogram determined by vectors $\vec{u} = \binom{1}{2}{3}$ and $\vec{v} = \binom{4}{5}{6}$.

The cross-product vector isolated in Example 1 equals $\binom{-3}{6}{-3}$.
Area equates universally to the numerical magnitude of this structural vector:
$\text{Area} = \sqrt{(-3)^2 + 6^2 + (-3)^2} = \sqrt{9 + 36 + 9} = \sqrt{54} \approx \mathbf{7.35}$.
The subsequent triangle partitioned identically equals $\frac{7.35}{2} \approx \mathbf{3.67}$.

EXAMPLE 4

Calculate the isolated bounded area for the triangle spanning corner nodes $A(1,1,1)$, $B(1,3,1)$, and $C(-3,3,4)$.

Determine two connected directional vectors spanning outwardly from a mutual anchor coordinate:
$\vec{AB} = \binom{0}{2}{0}$ and $\vec{AC} = \binom{-4}{2}{3}$.
Compile the structural cross product $\vec{AB} \times \vec{AC}$:
$\binom{(2)(3) - (0)(2)}{(0)(-4) - (0)(3)}{(0)(2) - (2)(-4)} = \binom{6}{0}{8}$.
The triangle area extracts to half the matrix output magnitude:
$\text{Area} = \frac{1}{2} \sqrt{6^2 + 0^2 + 8^2} = \frac{1}{2} \sqrt{100} = \mathbf{5}$.