Is magnitude of vector cross product always equal to dot product?

Question

Is magnitude of vector cross product always equal to dot product ?

What I mean to say is that

|vector cross product|(magnitude) = ABcos$\theta$ or magnitude of dot product. ^^ This statement up there means that magnitude of cross product = magnitude of dot product

is $$|A\times B|=|A||B|\sin(\theta)$$ $$A\cdot B=|A||B|\cos(\theta).$$

Answer 1

Your equation is incorrect. What is true is $$|A\times B|=|A||B|\sin(\theta)$$ $$A\cdot B=|A||B|\cos(\theta).$$

Answer 2

The magnitude of the cross product equals the dot product when $$|A| |B| \sin \theta = |A| |B| \cos \theta \\ \iff |A|=0, \ \text{or} \ |B|=0, \ \text{or} \ \sin\theta =\cos \theta \iff \theta =\frac{\pi}{4}$$

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The magnitude of the cross product equals the magnitude of the dot product when $$|A| |B| \sin \theta = |A| |B| | \cos \theta| \\ \iff |\tan \theta| =1 \iff \theta =\frac{\pi}{4} , \frac{3\pi}{4} $$

or if one of $|A|, |B|$ is zero.

So no, neither of these is always the case.

Here, it is assumed that $\theta \in [0, \pi)$.

Answer 3

Since$$\begin{align}(A\cdot B)^2+|A\times B|^2&=A_iB_iA_jB_j+\epsilon_{ijk}\epsilon_{ilm}A_jA_lB_kB_m\\&=(\delta_{jk}\delta_{lm}+\epsilon_{ijk}\epsilon_{ilm})A_jA_lB_kB_m\\&=\delta_{jl}\delta_{km}A_jA_lB_kB_m\\&=A^2B^2,\end{align}$$the correct result is $|A\times B|=\sqrt{A^2B^2-(AB\cos\theta)}=AB|\sin\theta|$.This equals $AB\cos\theta$ iff $\cos\theta=\tfrac{1}{\sqrt{2}}$, i.e. $A,\,B$ are separated by an angle of $\pi/4$.