1. ## normal vector

If $\displaystyle n$ is a normal vector of a plane, then it is perp. to every vector on the plane right? Let $\displaystyle \vec{r}, \vec{r_0}, \vec{r_1} -\vec {r_0}$ be vectors on the plane.

Then $\displaystyle n \cdot (\vec{r}-\vec{r_0}) = 0$.

Then shouldnt $\displaystyle n \cdot \vec{r_0} = n \cdot \vec{r_1} = 0$?

2. Originally Posted by shilz222
If $\displaystyle n$ is a normal vector of a plane, then it is perp. to every vector on the plane right? Let $\displaystyle \vec{r}, \vec{r_0}, \vec{r_1} -\vec {r_0}$ be vectors on the plane.

Then $\displaystyle n \cdot (\vec{r}-\vec{r_0}) = 0$.

Then shouldnt $\displaystyle n \cdot \vec{r_0} = n \cdot \vec{r_1} = 0$?
when we refer to $\displaystyle n \cdot (\vec{r}-\vec{r_0}) = 0$, we are referring to the (vector) equation of a plane. in this "construct", $\displaystyle r - r_0$ is a vector in the plane, but $\displaystyle r$ and $\displaystyle r_0$ are actually vectors beginning at the origin whose terminal points are points in the plane, so individually, they are not necessarily perpendicular to the normal vector