The Magnitude of the Cross Product of Two Vectors is Equal to the Area of the Parallelogram They Form
Theorem
The magnitude of the cross product of two vectors $\mathbf{A}$ and $\mathbf{B}$, when the angle between them is $\theta$, is as follows:
$$ \left| \mathbf{A}\times \mathbf{B}\right| =\left|\mathbf{A}\right|\left| \mathbf{B} \right|\sin \theta $$
And this is equal to the area of the parallelogram that the two vectors form.
Proof
Let’s say the two vectors $\mathbf{A}=(A_{x},A_{y},A_{z})$ and $\mathbf{B}=(B_{x},B_{y},B_{z})$ are as shown in the figure above. Then
- part 1. Area of the parallelogram
The area of the parallelogram is the product of base and height, so it is as follows.
$$ |\mathbf{A}| | \mathbf{B}|\sin\theta $$
- part 2. Magnitude of the cross product
$$ \begin{align*} \left|\mathbf{A}\times \mathbf{B}\right|^{2} &= \left| (A_{y}B_{z}-A_{z}B_{y})\hat{\mathbf{x}}+(A_{z}B_{x}-A_{x}B_{z})\hat{\mathbf{y}}+(A_{x}B_{y}-A_{y}B_{x})\hat{\mathbf{z}} \right|^{2} \\ &=(A_{y}B_{z}-A_{z}B_{y})^{2}+(A_{z}B_{x}-A_{x}B_{z})^{2}+(A_{x}B_{y}-A_{y}B_{x})^{2} \\ &= A_{y}^{2}B_{z}^{2}-2A_{y}A_{z}B_{y}B_{z}+A_{z}^{2}B_{y}^{2} \\ &\quad+ A_{z}^{2}B_{x}^{2}-2A_{z}A_{x}B_{z}B_{x}+A_{x}^{2}B_{z}^{2} \\ &\quad+ A_{x}^{2}B_{y}^{2}-2A_{x}A_{y}B_{x}B_{y}+A_{y}^{2}B_{x}^{2} \\ &\quad \color{red}{+A_{x}^{2}B_{x}^{2}+A_{y}^{2}B_{y}^{2}+A_{z}^{2}B_{z}^{2}} \color{blue}{-A_{x}^{2}B_{x}^{2}-A_{y}^{2}B_{y}^{2}-A_{z}^{2}B_{z}^{2}} \\ &= A_{x}^{2}(B_{x}^{2}+B_{y}^{2}+B_{z}^{2})+ A_{y}^{2}(B_{x}^{2}+B_{y}^{2}+B_{z}^{2})+ A_{z}^{2}(B_{x}^{2}+B_{y}^{2}+B_{z}^{2}) \\ &\quad -(A_{x}^{2}B_{x}^{2}+A_{y}^{2}B_{y}^{2}+A_{z}^{2}B_{z}^{2})^{2} \\ &= (A_{x}^{2}+A_{y}^{2}+A_{z}^{2})(B_{x}^{2}+B_{y}^{2}+B_{z}^{2})-(A_{x}^{2}B_{x}^{2}+A_{y}^{2}B_{y}^{2}+A_{z}^{2}B_{z}^{2})^{2} \\ &= |\mathbf{A}|^{2}|\mathbf{B}|^{2}-|\mathbf{A}\cdot\mathbf{B}|^{2} \\ &= |\mathbf{A}|^{2}|\mathbf{B}|^{2}-|\mathbf{A}|^{2}|\mathbf{B}|^{2}\cos ^{2 }\theta \\ &=|\mathbf{A}|^{2}|\mathbf{B}|^{2}(1-\cos ^{2 }\theta) \\ &=|\mathbf{A}|^{2}|\mathbf{B}|^{2}\sin ^{2 } \theta \end{align*} $$
Therefore,
$$ \left|\mathbf{A} \times \mathbf{B} \right|=\left|\mathbf{A}\right|\left| \mathbf{B} \right|\sin \theta $$
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