# Why is the product of a normal vector and a vector on the plane equal to the equation of the plane?

and basically it seems like the dot product of a normal vector to a plane and a vector on the plane is equal to the equation of the plane. What is the intuition behind this?

It seems like the equation of the plane is Ax + By + Cz = D

I was watching this video and it seems like you can define a plane just with a normal vector and a point on the plane. But how do you know how big this plane is?

• What do you mean by how big the plane is? Have you studied dot products?
– Karl
Jan 27, 2016 at 18:53
• The dot product of normal to the plane and any other vector is equal to the normal distance of the plane from the origin: if and only if the vector describes a point lying on that plane.
– ARi
Jan 27, 2016 at 18:55
• Have you seen $a \cdot b = |a||b| \cos \theta$ ?
– Karl
Jan 27, 2016 at 18:58
• yes I've seen that equation. It's law of cosines right? Jan 27, 2016 at 19:16
• The dot product is a way of measuring how perpendicular the vectors are. $\cos 90^{\circ} = 0$ forces the dot product to be zero. Ignoring the cases where the magnitude of the vectors is zero anyway.
– Karl
Jan 27, 2016 at 19:34

The linked reading isn't saying that the dot product is equal to the equation of the plane, it's saying that setting the dot product equal to 0 gives the equation of the plane. Following the notation of the linked page, let $\vec{n} = \langle a, b, c \rangle$ be the vector normal to the plane, let $\vec{r}_{0}$ be the position vector of a point in the plane $P_0 = (x_0, y_0, z_0)$, and let $\vec{r}$ be the position vector of an arbitrary point in the plane $P = (x, y, z)$. $a, b, c, x_0, y_0, z_0$ are all known; $x, y, z$ are free variables. The vector $\vec{r} - \vec{r}_{0}$ is in the plane, implying it's orthogonal to $\vec{n}$, thus $\vec{n} \cdot (\vec{r} - \vec{r}_{0}) = 0$. Substituting the component forms of the vectors into this equation gives us an equation in $x, y, z$ that defines the plane.