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

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 $$|A\times B|=|A||B|\sin(\theta)$$ $$A\cdot B=|A||B|\cos(\theta).$$ • Are yous sure about $\cos \theta?$
– mfl
Dec 29, 2020 at 18:36
• Yes.It is in my book as well. A.B = AB cos theta@mfl
– user864449
Dec 29, 2020 at 18:41
• Cross product magnitude has sin, not cos. Dot product has cos. Example two parallel vectors, cross prod. $=0$. Dec 29, 2020 at 18:41
• @herbsteinberg Typo
– user864449
Dec 29, 2020 at 18:43

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

• $|A\times B|=A\cdot B$ if and only if $\sin(\theta)=\cos(\theta)$, which is if $\theta=45^\circ$. Dec 29, 2020 at 18:45
• When can they be equal.There must be some condition
– user864449
Dec 29, 2020 at 18:45
• For the cross product, you might want $| \sin \theta |$. Dec 29, 2020 at 18:45
• @peterag It's common to assume $\theta\in[0,180]$ since we are viewing $\theta$ geometrically as the angle between the vectors. So the absolute value is not necessary. Dec 29, 2020 at 18:47
• Hence the 'might'... but I did think you should comment. Dec 29, 2020 at 18:47

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

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 $$\theta \in [0, \pi)$$.

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 $$\pi/4$$.

• Please check my edit sir
– user864449
Dec 29, 2020 at 18:49
• @user282657 You're still putting $\cos\theta$ on both sides, when one should use $\sin\theta$.
– J.G.
Dec 29, 2020 at 19:24