Suppose we have the plane with equation $3x-7z=12$. How to find its normal vector?
The plane with equation $Ax+By+Cz+D=0$ has the normal vector $\mathbb{n}=(A,B,C)$.
Using this we get that above plane has normal vector $(3,0,-7)$, right?
Let's apply another method. Take three points which lies on the plane, namely $A=(4,0,0), \ B=(0,0,-\frac{12}{7}), \ C=(1,0,\frac{-9}{7})$ then vector $AB=(-4,0,-\frac{12}{7})$ and $AC=(-3,0,-\frac{9}{7})$. Taking their cross product we get zero vector.
What is wrong in my idea?
Best Answer
In response to the first part:
Suppose two points, $P(x,y,z)$ and $P_0(x_0, y_0, z_0)$ lie on a plane with a normal vector $\mathbf{n}$. Also, let $\mathbf{r}$ denote the position vector of $P$ and $\mathbf{r_0}$ denote the position vector of $P_0$.
The picture clearly shows that $\mathbf{n}\cdot(\mathbf{r-r_0}) = 0$, since the two vectors are perpendicular. Then, $$\mathbf{n}\cdot \mathbf{r} = \mathbf{n}\cdot\mathbf{r_0}$$ In your case, we have $3x+0y-7z = 12$, which is equivalent to $(3,0,-7)\cdot(x,y,z) = 12$. Comparing with above, we have the components of the normal to the plane are $(3,0,-7)$.
Second question:
You have made a computation mistake, as has been pointed out in the other posts. Your idea is nonetheless true: if you have two direction vectors lying in the plane, then their cross product will result in a vector orthogonal to both these vectors, i.e. normal to the plane.