Explanation of: The vector $(\alpha, \beta)$ is parallel to the line $Ax + By + C = 0$ if and only if $A\alpha +B\beta = 0$ **I just stumbled on this passus in my textbook and I cannot really make sense of it:** > The vector $(\alpha, \beta)$ is parallel to the line $Ax + By + C = 0$ if and only if $A\alpha +B\beta = 0$. In other words, the vector $\vec{n} = (A,B)$ is perpendicular to this line and is called the _normal vector_ to the line. **My analysis:** First of all the statement $A\alpha +B\beta = 0$ represent the scalar (or dot) product of the two vectors, i.e $(\alpha, \beta) \cdot (A, B) = A\alpha +B\beta$. This is $0$ if the vectors are perpendicular and $1, -1$ if the vectors are parallel. Why does it say in my textbook that the The vector $(\alpha, \beta)$ is parallel to the line $Ax + By + C = 0$ if and only if $A\alpha +B\beta = 0$? Shouldn't it be perpendicular instead? Please explain to me if there is something that I have missed? Thank you!

If $(A,B)$ is normal (perpendicular) to the line, then any vector perpendicular to $(A,B)$ is parallel to the line, and this is achieved by setting the scalar product to zero.

Another cruder way of seeing this is by some calculation as follows:

Suppose first that $B\
eq 0$ so that $y=-\frac AB x-\frac CB$

The gradient of this line is $-\frac AB$

The equation of the line through the origin having this gradient is $y=-\frac AB x$ and this is parallel to the original line.

This equation can be rewritten $Ax+By=0$ and $(\alpha,\beta)$ lies on the line when $A\alpha+B\beta=0$. This formulation also works when $B=0$ as you can check.