Cross Product and Triple Product

For vectors $\vec{a}, \vec{b}, \vec{c}$ and scalar $k$:

• Anti-commutative: $\vec{a}\times\vec{b} = -\vec{b}\times\vec{a}$
• Distributive: $\vec{a}\times(\vec{b}+\vec{c}) = \vec{a}\times\vec{b} + \vec{a}\times\vec{c}$
• Scalar: $(k\vec{a})\times\vec{b} = k(\vec{a}\times\vec{b})$
• Self: $\vec{a}\times\vec{a} = \vec{0}$
• Parallel test: $\vec{a}\times\vec{b} = \vec{0} \iff \vec{a}\,\|\,\vec{b}$ (or one is zero)
• Not associative: $(\vec{a}\times\vec{b})\times\vec{c} \neq \vec{a}\times(\vec{b}\times\vec{c})$ in general
• Basis sums (compute in components once, then quote): $|\vec{a}\times\hat{i}|^2 + |\vec{a}\times\hat{j}|^2 + |\vec{a}\times\hat{k}|^2 = 2|\vec{a}|^2$ — each cross product drops one component, so every $a_i^2$ is counted twice; and $(\vec{r}\cdot\hat{i})(\vec{r}\times\hat{i}) + (\vec{r}\cdot\hat{j})(\vec{r}\times\hat{j}) + (\vec{r}\cdot\hat{k})(\vec{r}\times\hat{k}) = \vec{0}$, since it equals $\vec{r}\times\vec{r}$ by distributivity. Their dot-product cousin is $(\vec{a}\cdot\hat{i})^2+(\vec{a}\cdot\hat{j})^2+(\vec{a}\cdot\hat{k})^2 = |\vec{a}|^2$.
• Type rules — which expressions even make sense: a dot product needs two VECTORS and returns a scalar; a cross product needs two vectors and returns a vector. So $(\vec{a}\cdot\vec{b})\cdot\vec{c}$, $(\vec{a}\cdot\vec{b})\times\vec{c}$, $\vec{a}\times(\vec{b}\cdot\vec{c})$ and $(\vec{a}\cdot\vec{b})\cdot(\vec{c}\cdot\vec{d})$ as a dot are all meaningless (a scalar is being dotted or crossed); $(\vec{a}\cdot\vec{b})\,\vec{c}$ (scalar times vector), $(\vec{a}\times\vec{b})\cdot\vec{c}$ and $(\vec{a}\times\vec{b})\times\vec{c}$ are fine. "How many of these are not defined?" is answered by counting the scalars fed into a $\cdot$ or $\times$.

For non-zero $\vec{a}, \vec{b}$ at angle $\theta$: $|\vec{a}\times\vec{b}| = |\vec{a}|\,|\vec{b}|\sin\theta$. Area of parallelogram with sides $\vec{a}, \vec{b}$ is $|\vec{a}\times\vec{b}|$; area of triangle with the same sides is $\tfrac{1}{2}|\vec{a}\times\vec{b}|$. Lagrange identity: $|\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2\,|\vec{b}|^2$.

If $\vec{a}, \vec{b}$ are not parallel, a unit vector perpendicular to both is $\hat{n} = \pm\dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}$. Both signs are valid answers unless the question specifies a direction (right-hand rule, towards a third vector, etc.).

If a force $\vec{F}$ acts at a point $P$ and we measure its moment about a point $O$, then $\vec{M} = \overrightarrow{OP} \times \vec{F}$. Moment magnitude is $|\vec{M}| = |\overrightarrow{OP}||\vec{F}|\sin\theta = d\,|\vec{F}|$ where $d$ is the perpendicular distance from $O$ to the line of action.

For $\vec{a}, \vec{b}, \vec{c}$ in $\mathbb{R}^3$, the scalar triple product is $[\vec{a}\,\vec{b}\,\vec{c}] = \vec{a}\cdot(\vec{b}\times\vec{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$. Coplanarity criterion: $\vec{a}, \vec{b}, \vec{c}$ are coplanar iff $[\vec{a}\,\vec{b}\,\vec{c}] = 0$. Volume of the parallelepiped on the three vectors is $|[\vec{a}\,\vec{b}\,\vec{c}]|$.

Cyclic identity: $\vec{a}\cdot(\vec{b}\times\vec{c}) = \vec{b}\cdot(\vec{c}\times\vec{a}) = \vec{c}\cdot(\vec{a}\times\vec{b})$. Anti-cyclic: swapping any two negates the value, e.g. $\vec{a}\cdot(\vec{c}\times\vec{b}) = -\vec{a}\cdot(\vec{b}\times\vec{c})$. Linearity in each slot: $[(\alpha\vec{u}+\beta\vec{v})\,\vec{b}\,\vec{c}] = \alpha[\vec{u}\,\vec{b}\,\vec{c}] + \beta[\vec{v}\,\vec{b}\,\vec{c}]$. If $\vec{a}, \vec{b}, \vec{c}$ are coplanar, $[\vec{a}\,\vec{b}\,\vec{c}] = 0$ and the derived cyclic sum collapses to zero too.

For any $\vec{a}, \vec{b}, \vec{c} \in \mathbb{R}^3$: $\vec{a}\times(\vec{b}\times\vec{c}) = (\vec{a}\cdot\vec{c})\,\vec{b} - (\vec{a}\cdot\vec{b})\,\vec{c}$. By anti-commutativity, $(\vec{a}\times\vec{b})\times\vec{c} = -\,\vec{c}\times(\vec{a}\times\vec{b}) = (\vec{a}\cdot\vec{c})\,\vec{b} - (\vec{b}\cdot\vec{c})\,\vec{a}$. Either form lies in the plane of the two inner vectors and is perpendicular to the outer one.