How to find a normal vector given a plane with the following equation: $4(−8)−14(−3)+6=0$

Question

How would I find a vector normal $$ to the plane with the equation:

$4(−8)−14(−3)+6=0$.

So I first distribute:

$4x-32-14y+42+6z=0$

then I combine like terms and move it to the other side:

$4x-14y+6z=-10$

So my answer for this normal vector is:

$\left\langle-32,42,0\right\rangle$

But it doesn't seem to be the right answer. What's my problem here? Thanks in advance.

Answer 1

A normal vector to a plane $ax+by+cz=d$ is $(a,b,c)$. So, in your case an answer would be $(4,-14,6)$. Of course, $(2,-7,3)$ is also a correct answer.

Answer 2

Suppose there’s a point $A(x_0,y_0,z_0)$ on a plane $E$ , whose normal vector $\overrightarrow {n}=(a,b,c)$.

Let $P(x,y,z)$ be a point on $E$ and is different from $A$.

Since $\overrightarrow{AP}$ is a vector on $E$, $\overrightarrow {n}$ is perpendicular with $\overrightarrow{AP}$. Therefore: $$\overrightarrow{n} \cdot \overrightarrow{AP}=(a,b,c)\cdot(x-x_0,y-y_0,z-z_0)=0$$

And you get:

$$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$$

And this is the equation representing the plane $E$ because every point $P(x,y,z)$ on $E$ satisfies this equation.

This is why $(a,b,c)$ is the normal vector of the plane: $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$. I hope this helps you know the reason behind this question better.