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The Magnitude of the Cross Product of Two Vectors is Equal to the Area of the Parallelogram They Form 📂Mathematical Physics

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 A\mathbf{A} and B\mathbf{B}, when the angle between them is θ\theta, is as follows:

∣A×B∣=∣A∣∣B∣sin⁡θ \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

5F17E9AC2.png

Let’s say the two vectors A=(Ax,Ay,Az)\mathbf{A}=(A_{x},A_{y},A_{z}) and B=(Bx,By,Bz)\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.

∣A∣∣B∣sin⁡θ |\mathbf{A}| | \mathbf{B}|\sin\theta

  • part 2. Magnitude of the cross product

∣A×B∣2=∣(AyBz−AzBy)x^+(AzBx−AxBz)y^+(AxBy−AyBx)z^∣2=(AyBz−AzBy)2+(AzBx−AxBz)2+(AxBy−AyBx)2=Ay2Bz2−2AyAzByBz+Az2By2+Az2Bx2−2AzAxBzBx+Ax2Bz2+Ax2By2−2AxAyBxBy+Ay2Bx2+Ax2Bx2+Ay2By2+Az2Bz2−Ax2Bx2−Ay2By2−Az2Bz2=Ax2(Bx2+By2+Bz2)+Ay2(Bx2+By2+Bz2)+Az2(Bx2+By2+Bz2)−(Ax2Bx2+Ay2By2+Az2Bz2)2=(Ax2+Ay2+Az2)(Bx2+By2+Bz2)−(Ax2Bx2+Ay2By2+Az2Bz2)2=∣A∣2∣B∣2−∣A⋅B∣2=∣A∣2∣B∣2−∣A∣2∣B∣2cos⁡2θ=∣A∣2∣B∣2(1−cos⁡2θ)=∣A∣2∣B∣2sin⁡2θ \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,

∣A×B∣=∣A∣∣B∣sin⁡θ \left|\mathbf{A} \times \mathbf{B} \right|=\left|\mathbf{A}\right|\left| \mathbf{B} \right|\sin \theta

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