find the equation of line given normal direction and normal distance

Question

How can I find the equation of a line given only the normal direction and normal distance to this line?

where a is the normal direction, d is the normal distance and O is the origin.

Answer 1

The vector $(\cos a, \sin a)$ is normal to the line, which passes through the point $(x_0,y_0)=(d\cos a, d\sin a)$, hence the equation of the line is $$\color{red}{x\cos a+y \sin a}=x_0\cos a+y_0 \sin a=d(\cos^2a+\sin^2a)\color{red}{=d}.$$

Answer 2

In $x-y$ plane the equation of the line is given by

$$ax+by=d$$

beeing the normal $n=(a,b)$ such that $|n|=1$, indeed by dot product between $OP$ and $n$ for any point $P=(x,y)$ on the line we have

$$(x,y)\cdot (a,b)=ax+by=|OP|\cos \theta=d$$

Answer 3

You can see that the vector $\overrightarrow{OB}= (d\cos a , d\sin a )$ is orthogonal to the line. If you consider other point P(x, y) in the line, then: $$\overrightarrow{OP}\cdot \overrightarrow{OB} = (x, y)\cdot (d\cos a, d\sin a) = xd \cos a + yd\sin a = d|\overrightarrow{OP}|\cos\theta.$$

Observe than $\cos \theta$ is always equal to $d/|\overrightarrow{OP}|$. Therefore it is: $$ xd \cos a + yd\sin a = d|\overrightarrow{OP}|\cos\theta = d^2.$$ That is to say, that the equation of the line is: $$x \cos a + y\sin a = d.$$

Good Luck!