Properties of Vector Product

(i) Anti-commutative (not commutative). By definition, $\vec b\times\vec a=|\vec b||\vec a|\sin\theta\,(-\hat n)$, since $\vec b,\vec a,-\hat n$ (rather than $\vec b,\vec a,\hat n$) form a right-handed system. So $\vec b\times\vec a=-(\vec a\times\vec b).$ The vector product is genuinely non-commutative.

(ii)–(iii) The zero-product test. If $\vec a,\vec b$ are collinear (parallel), $\theta=0$ or $\pi$, so $\sin\theta=0$ and $\vec a\times\vec b=\vec 0$. Conversely, $\vec a\times\vec b=\vec 0\iff \vec a=\vec0$ or $\vec b=\vec0$ or $\vec a\parallel\vec b$. So for non-zero vectors, $\vec a\times\vec b=\vec 0$ is exactly the condition $\vec a\parallel\vec b$ — the mirror image of the dot-product's perpendicularity test. In particular, $\vec a\times\vec a=\vec 0$ always.

(iv) The axis unit vectors. $\hat i\times\hat i=\hat j\times\hat j=\hat k\times\hat k=\vec 0$ (each parallel to itself). Cyclically (right-handed system): $\hat i\times\hat j=\hat k,\qquad \hat j\times\hat k=\hat i,\qquad \hat k\times\hat i=\hat j,$ and reversing any pair flips the sign: $\hat j\times\hat i=-\hat k,\ \hat k\times\hat j=-\hat i,\ \hat i\times\hat k=-\hat j$.

(v) Perpendicular vectors. If $\theta=\pi/2$, then $|\vec a\times\vec b|=|\vec a||\vec b|$ (since $\sin\frac\pi2=1$) — the maximum possible magnitude for given $|\vec a|,|\vec b|$.

(vi) Scalars pull straight out. For scalars $m,n$: $m\vec a\times n\vec b=(mn)(\vec a\times\vec b)=m(\vec a\times n\vec b)=n(m\vec a\times\vec b).$

(vii) Distributive. $\vec a\times(\vec b+\vec c)=\vec a\times\vec b+\vec a\times\vec c$, and $(\vec a+\vec b)\times\vec c=\vec a\times\vec c+\vec b\times\vec c$; this extends to subtraction, $\vec a\times(\vec b-\vec c)=\vec a\times\vec b-\vec a\times\vec c$, and to sums of any number of vectors, e.g. $\vec a\times(\vec b+\vec c+\vec d)=\vec a\times\vec b+\vec a\times\vec c+\vec a\times\vec d$.

(viii) Coordinate (determinant) formula. For $\vec a=a_1\hat i+a_2\hat j+a_3\hat k,\ \vec b=b_1\hat i+b_2\hat j+b_3\hat k$, expand $\vec a\times\vec b$ term by term using (iv), and collect: $\vec a\times\vec b=(a_2b_3-a_3b_2)\hat i-(a_1b_3-a_3b_1)\hat j+(a_1b_2-a_2b_1)\hat k=\begin{vmatrix}\hat i&\hat j&\hat k\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix}.$ …