# How do you derive a normal vector from the equation of a line?

I've seen that to find a normal vector to a line such as $3x+4y-1=0$ people take the coefficients and say $(3,4)$ is a normal vector?

Why does this work? How are the coefficients related to the normal?

• @ZacharySelk I think its a plane in 3 dimensions. Although normal vector will be <3,4,0> right? Commented Jan 8, 2018 at 4:18
• @prog_SAHIL They said it's a line but if it was a plane you would be correct.
– user223391
Commented Jan 8, 2018 at 4:21
• @ZacharySelk I've corrected it
– K.M.
Commented Jan 8, 2018 at 4:22
• The coefficients, not the constant term though, form a normal vector. Commented Jan 8, 2018 at 4:28
• Trivially. No pun intended. Commented Jan 8, 2018 at 10:45

Two vectors $a,b$ are normal iff $a\cdot b=a_1b_1+...+a_2b_2=0$.

If you have a vector, $(x,y)$ and you want to find vectors that are normal to it you want to find a vector $(a,b)$ such that $ax+by=0$. So a normal vector to the line $3x+4y=0$ is simply $(3,4)$.

If you have a constant on the right side, it just moves the line up or down. It doesn't change anything else. So a normal vector will still be $(3,4)$.

• The normal vector is not unique, so you might want to call this "a normal vector"? Commented Jan 8, 2018 at 4:29
• @max_zorn sure.
– user223391
Commented Jan 8, 2018 at 4:30
• sorry I don't know If I'm being retarded but I don't get how (3,4) is a normal vector to the line 3x+4y=0
– K.M.
Commented Jan 8, 2018 at 4:35
• @K.M. the vector $(3,4)$ is orthogonal to any vector $(x,y)$ which is on that line because $$(3,4)\cdot (x,y)=3x+4y=0$$
– Dave
Commented Jan 8, 2018 at 5:20
• @K.M. To be more specific, $(3,4)$ is normal to any vector pointing along the line $3x + 4y + c = 0$ for any constant $c$. If $(x_1, y_1)$ and $(x_2, y_2)$ are two points on the line, then $(x_2 - x_1, y_2 - y_1)$ is a vector pointing along the line. Subtracting $3x_1 + 4y_1 + c = 0$ from $3x_2 + 4y_1 + c = 0$, you get $$3(x_2 - x_1) + 4(y_2 - y_1) = 0$$which tells you that $(3,4)$ is normal to the vector $(x_2 - x_1, y_2 - y_1)$. Commented Jan 8, 2018 at 14:39

If the line equation is $$ax+by+c=0$$

then, the normal vector is $\vec{n}=\left(\begin{array}{c}a \\ b\end{array}\right)$, and the direction vector is $\vec{v}=\left(\begin{array}{c}-b \\ a\end{array}\right)$

Demonstration:

First, we begin by showing that $\vec{n}=\left(\begin{array}{c}a \\ b\end{array}\right)$

It is easy to see that if $a=(x_a,y_a)$, $b=(x_b,y_b)$ are two points from the given line then, $$\vec{u}=\left(\begin{array}{c}x_b-x_a \\ y_b-y_a \end{array}\right)$$ is a direction vector of the line.

then the scalar product of $n$ et $u$ must be $0$ to say that $n$ is indeed a normal vector. $$\vec{n}\cdot \vec{u}=a(x_b-x_a)+b(y_b-y_a)=ax_b-ax_a+by_b-by_a=c-c=0$$

Then to show that $\vec{v}=\left(\begin{array}{c}-b \\ a\end{array}\right)$ is a direction vector for the line all we have to do is to calculate the scalar product of $\vec{n}$ and $\vec{v}$. $$\vec{n}\cdot \vec{v}=-ab+ab=0$$ So the vector $\vec{v}$ is orthogonal to the vector $\vec{n}$ which is a normal vector, hence $\vec{v}$ is a direction vector.

• Good answer. Maybe it is not the best idea to use the same variable name for the coefficients and points on the line. This might confuse. Commented Jan 8, 2018 at 8:14
• thanks you for the remark, i hope it is less confusing now. Commented Jan 8, 2018 at 11:14

A vector parallel to the line is obtained by joining two points, say

$$\vec p=(x_1,y_1)-(x_0,y_0).$$

Let $\vec n:=(3,4)$. Then by the equation of the line ($3x+4y=1$),

$$\vec n\cdot\vec p=(3x_1+4y_1)-(3x_0+4y_0)=1-1=0$$

which shows that $\vec n\perp\vec p$.

Two lines are perpendicular if their slopes are opposite reciprocal. The slope of your line is $-3/4$ so you are looking for a line with slope of $4/3$. If you let $x=3t$, $y=4t$ the slope will be 4/3 and you are done.

• whats t here? putting 3t inplace of x and 4t in place of y doesnt make sense (3t+4t-1=0 is 7t-1=0)
– A.B
Commented Sep 19, 2020 at 19:03

$$3x+4y=0,$$ or $$\vec n\cdot\vec p=0.$$
Completely analogous to the case of a plane. .. $ax+by+cz=d$ has normal vector $(a,b,c)$... this is because the dot product $(x-x_0, y-y_0, z-z_0)\cdot (a,b,c)=0$, where $d=ax_0+by_0+cz_0$, for any point $(x_0, y_0, z_0)$ on the plane. ..