 Solution

LetThe vector triple product identity $(\mathbf{a}\wedge \mathbf{b})\wedge \mathbf{a}=|\mathbf{a}{|}^2\mathbf{b}-(\mathbf{a}\cdot \mathbf{b})\mathbf{a}$ shows that ${\mathbf{r}}_0\wedge \mathbf{a}=\mathbf{b}$. Moreover, $(\mathbf{r}-{\mathbf{r}}_0)\wedge \mathbf{a}=0$ exactly when $\mathbf{r}-{\mathbf{r}}_0$ is a scalar multiple of $\mathbf{a}$. Thus the given line isAlso $\mathbf{a}\wedge {\mathbf{r}}_0=-\mathbf{b}$, so $\mathbf{N}=\mathbf{a}\wedge (\mathbf{y}-{\mathbf{r}}_0)$. The assumption $\mathbf{N}\ne 0$ says precisely that $\mathbf{y}$ is outside $L$. Consequently $L$ and $\mathbf{y}$ determine a unique plane, with normal vector $\mathbf{N}$.

To identify its equation, use cyclic invariance of the scalar triple product:For $\mathbf{r}\in L$, this equals $\mathbf{y}\cdot \mathbf{b}$, while $\mathbf{r}\cdot \mathbf{b}=\mathbf{r}\cdot (\mathbf{r}\wedge \mathbf{a})=0$. Hence $\mathbf{r}\cdot \mathbf{N}=\mathbf{y}\cdot \mathbf{b}$ throughout $L$. The point $\mathbf{y}$ also satisfies that equation because $\mathbf{y}\cdot (\mathbf{a}\wedge \mathbf{y})=0$. Since $\mathbf{N}\ne 0$, the required plane is therefore exactlyThis also proves the reusable plane through a point and a line given by a cross product construction.