1 row matrix determinant

Question

In Line-plane intersection in parametric form article determinant of matrix is used to check whether solution exists. Matrix elements are vectors. I don't understand how determinant is found and why it results in dot product of first element with cross product of 2 other elements. As far as I concerned, we can calculate determinant for square matrix only. I appreciate any feedback. Thank you and have a nice day!

Answer 1

First, notice that at the very start of the article, the vectors $\mathbf{1}_a$ and $\mathbf{1}_b$ are defined in the form $\mathbf{1}_j:=(x_j,y_j,z_j)\in\mathbb R^3$. So, these are really points in space. when taking the determinant $$\det(\begin{bmatrix} -\mathbf{1}_{ab} & \mathbf{p}_{01} & \mathbf{p}_{02} \end{bmatrix})\,,$$ we ought to interpret the matrix $\begin{bmatrix} -\mathbf{1}_{ab} & \mathbf{p}_{01} & \mathbf{p}_{02} \end{bmatrix}$ as a "block" matrix with three entries, each of which is a $3\times 1$ column-matrix. Thus, in this convention, the matrix $\begin{bmatrix} -\mathbf{1}_{ab} & \mathbf{p}_{01} & \mathbf{p}_{02} \end{bmatrix}$ is really a $3\times3$ square matrix as desired. This is why the determinant of such a matrix makes sense.

The fact that $$ \det(\begin{bmatrix} -\mathbf{1}_{ab} & \mathbf{p}_{01} & \mathbf{p}_{02} \end{bmatrix}) = -\mathbf{1}_{ab}\cdot(\mathbf{p}_{01}\times\mathbf{p}_{02}) $$ is really an application of the definition of a $3\times 3$ determinant. You should check this result by expanding along the first column of $\begin{bmatrix} -\mathbf{1}_{ab} & \mathbf{p}_{01} & \mathbf{p}_{02} \end{bmatrix}$, which in this case is $-\mathbf{1}_{ab}$. Can you check this by yourself ?