Cross product ($\times$)

The cross product $a\times b$ of two three-dimensional vectors is perpendicular to both, has magnitude $|a||b|\sin \theta$, and is oriented by the right-hand rule.

For a unit vector $\mathbf{n}$, real $\alpha \ne 0$, and real $\beta$, the displayed linear equation has the unique vector solutionTo prove it, split vectors into components parallel and orthogonal to $\mathbf{n}$. On the parallel line, the linear map is multiplication by $\alpha$. On the perpendicular plane, the operator $N\mathbf{v}=\mathbf{n}\times \mathbf{v}$ obeys $N^2=-I$, by the vector triple product. Therefore $(\alpha I-\beta N)(\alpha I+\beta N)=({\alpha }^2+{\beta }^2)I$ on that plane. Combining its inverse with the parallel solution gives the formula. This avoids component calculations in equations with an axial cross product term.

For any real three-dimensional vectors $a,b$, the displayed linear equation has the unique solutionTaking a dot product with $b$ first yields $(1+|b{|}^2)(b\cdot x)=a\cdot b$. Taking a cross product with $b$ and using the vector triple product identity givesSubstitution into the original equation cancels the longitudinal terms and proves the formula. Conversely direct substitution verifies it for all $b$, including $b=0$, since the denominator is positive. If $C_{b}x=b\times x$, the underlying identity for linear maps is $(I-C_b)(I+C_b+b{b}^T)=(1+|b{|}^2)I$, using $C_b^2=b{b}^T-|b{|}^{2}I$.

For a fixed unit vector $g\in {\mathbb{R}}^3$, the linear map $r\mapsto r\times g$ has kernel $\mathbb{R}g$ and image the perpendicular plane $g^{\perp }$. Thus $r\times g=f$ is solvable exactly when $f\cdot g=0$, and then its full solution set isFor the proof, a cross product with $g$ must be perpendicular to $g$. Conversely the vector triple product identity gives $g\times (r\times g)=r-(r\cdot g)g$, which determines the perpendicular orthogonal projection of $r$. The remaining parallel component is arbitrary. Finally $(g\times f)\times g=f$ when $f\cdot g=0$, verifying sufficiency.

Geometrically, the solution set is a straight line with direction $g$. Its nearest point to the origin is $g\times f$ and its distance from the origin is $|f|$. For a general nonzero fixed vector $a$, the corresponding formula is $r=(a\times f)/|a{|}^2+ta$, again subject to $f\cdot a=0$.