# Definition of Vector (Cross) Product

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.)

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.

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.

• 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.) Oct 1, 2020 at 11:35
• @JohnHughes Do you mean the use here of the scalar triple product?
– J.G.
Oct 1, 2020 at 17:19
• Yep --- that's exactly what I was thinking of. Thanks! Oct 1, 2020 at 17:31