Show that the $\det(M) = \det \begin{bmatrix} x & y & z \\ a_1& b_1 & c_1 \\ a_2& b_2 & c_2 \\ \end{bmatrix}=0$ is a equation of the line joining two distinct points $[a_1:b_1:c_1], [a_2: b_2: c_2] \in \mathbb{P}^2$.
I've just started studying projective geometry, and this is the first problem in the book "Notes On Geometry" by E.Rees.
Since it is so different from Euclidean Geometry, it takes a long time to digest Projective Geometry at this point, so I would like to get some help with this problem which is probably a simple one for ones who knows projective geometry.
$\det (M) = x(b_1c_2-c_1b_2)-y(a_1c_2-c_1a_2)+z(a_1b_2-b_1a_2)=0 $
And, the line joining $[a_1:b_1:c_1], [a_2: b_2: c_2] \in \mathbb{P}^2$ would be set of lines joining $(\lambda a_1, \lambda b_1, \lambda c_1)$, and $(\mu a_2, \mu b_2, \mu c_2)$ with $\lambda, \mu \in \mathbb{R} \setminus \{0\}$.
But, I can't see a clear relationship between them.
How can they be equal?
Understanding this clearly would really help a lot studying Projective Geometry, so please help.
$\det \begin{bmatrix} x & y & z \\ a_1& b_1 & c_1 \\ a_2& b_2 & c_2 \\ \end{bmatrix}=0$ if and only if the vectors $(x,y,z)$, $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ do not span the entire three-dimensional vector space. This is a key fact from linear algebra.
Therefore, whenever $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ are not on the same line (by "line" I mean a line through the origin in the vector space),
the equation says nothing more than that $(x,y,z)$ is on the same plane in the vector space as $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$.
Now you can re-express this statement projectively.