How can I find normal vector on the given line. For example if I have a line $3x – 5y = 1$, what would be the normal vector of this line? I am not sure whether it's useful or not, but we have one more random point on the line perpendicular to the given line, $P = (3,-2)$.
Point $P$ is just a random point onto the other line, not a intersecting point.
Best Answer
A normal vector of a 2-dimensional line will have the direction vector of an orthogonal line to it.
$$L: 3x-5y=1 \iff y = \dfrac{3}{5}x-\dfrac{1}{5}$$
So any line having slope $-\dfrac{5}{3}$ will be orthogonal to $L$. In other words, all normal vectors to $L$ will be a non-zero multiple of $\langle3,-5\rangle$. In general, a normal to
$$ax+by=c$$
will be a scalar multiple of $\langle a,b\rangle$.
If you're looking for the unique orthogonal line passing through your given $P$ then you use the point in the point-slope equation of a line.
$$y-(-2) = \dfrac{-5}{3}(x-3)$$