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I was given the following definition of the cross product:

The vector product $\underline{a}\times\underline{b}$ is defined as the vector with magnitude $\lvert\underline{a} \times \underline{b}\rvert = \vert\underline{a}\rvert\lvert\underline{b}\rvert \sin{\theta}$ and direction perpendicular to both $\underline{a}$ and $\underline{b}$, with $\theta$ the angle measured from $\underline{a}$ to $\underline{b}$

My understanding is that we measure angles anticlockwise by convention. And so if we were to try and compute $\underline{\hat{j}}\times\underline{\hat{i}}$ for example, the angle between these two vectors would be $\frac{3\pi}{2}$ and thus by the above definition, we have that the magnitude of $\underline{\hat{j}}\times\underline{\hat{i}}$ is $-1$ which is impossible because magnitudes of vectors must be non-negative.

I know I'm going wrong with my understanding here somewhere, I just don't understand where specifically.

(Also, I do understand right hand convention, and the fact that $\underline{\hat{j}}\times\underline{\hat{i}} = -\underline{\hat{i}}\times\underline{\hat{j}}$. However in this case, it is not that the magnitude is opposite, it's the direction which is opposite.)

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The quoted definition is careless: $|a\times b|=|a||b|(\sin\theta)c$ where $c\cdot c=1$ and $a,\,b,\,c$ form a right-handed system. The result is parallel to $c$ if the sine is positive, antiparallel to $c$ if the sine is negative, and the zero vector if the sine is zero.

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The cross product can also be defined as a mapping of $\vec a$ and $\vec b$ to a function $\vec x\mapsto \det(\vec a,\vec b, \vec x)$.

This function has properties that make it a homomorphism from the vector space to the underlying field (usually $\Bbb R$ or $\Bbb C$).

For finite dimensional vector spaces, there is a bijection (actually, another homomorphism) that maps these functions to the vector space, this identifying them with vectors (as John Hughes pointed out, co-vectors).

This kind of view of the cross product also opens up ways to generalize it to higher dimensions than just 3D.

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  • $\begingroup$ Right --- the cross product of vectors $a$ and $b$ is a co-vector (an element of the dual space of \Bbb R^3). This is very often the ideal way to think of things. In some fields, where your 3-space has a natural basis and inner product, the duality map $v \mapsto (u \mapsto (v \cdot u))$ gives a good way to identify this covector with a vector. (As I recall, this comes up in crystallography, for example...but I could be mis-remebering.) $\endgroup$ – John Hughes Oct 1 '20 at 11:35
  • $\begingroup$ @JohnHughes Do you mean the use here of the scalar triple product? $\endgroup$ – J.G. Oct 1 '20 at 17:19
  • $\begingroup$ Yep --- that's exactly what I was thinking of. Thanks! $\endgroup$ – John Hughes Oct 1 '20 at 17:31

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