# 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 $B\neq 0$ 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.

• Thank you for your answer! So my error was partly by assuming that $A\alpha +B\beta = 0$ represent the dot product of the vectors: $(\alpha, \beta) \cdot (A, B) = A\alpha +B\beta$? – Lukas Arvidsson Feb 21 '14 at 10:37
• @LukasArvidsson - you can use the dot product idea - zero dot product means the vectors are perpendicular. So we have $(A,B)$ perpendicular to the original line, and $(\alpha,\beta)$ perpendicular to $(A,B)$ hence parallel to the line. – Mark Bennet Feb 21 '14 at 10:40
• Thank you very much for your comment: What I have trouble understanding is this (but if it is correct I just need to study more :)):The vector $(\alpha, \beta)$ is parallel to the line $Ax + By + C = 0$ if and only if $A\alpha+B\beta=0$. – Lukas Arvidsson Feb 21 '14 at 10:46
• @LukasArvidsson Can I suggest you pick some values of $A, B \text{ and } C$ and draw a diagram containing the line, the vector $(A,B)$ from the origin and a vector perpendicular to $(A,B)$ at the origin. Then you may see what is going on. – Mark Bennet Feb 21 '14 at 10:54
• I will try that! Thank you! – Lukas Arvidsson Feb 21 '14 at 11:00