I believe that if you take the normal of those two vectors, it gives you the coefficients of that sought plane, but I don't understand why that is so. So you have two vectors on the same plane, the normal should give you a vector that's perpendicular to both. Using the hand notation to find the orientation of the vector, I don't understand how a vector perpendicular to those two vectors will give me the needed coefficients. When I visualize the plane formed by that vector, it's on the wrong plane altogether.
My question in regards to you can use normals to find planes came up in the following question:
Given two curves on a surface and a point, (2,1,3), find a plane tangent to that point. After finding the tangent vector corresponding to those curves and point, I believe you're supposed to use the normal to find the tangent plane, and I'm fine with the computations, but I don't understand why the normal of the two tangent vectors will give you the coefficients of the tangent plane.
Best Answer
The reason why the normal vector will give you the coefficients of the equation of the tangent plane is as follows:
Let the normal vector be $$\underline{n}=\left(\begin{matrix}a\\b\\c\end{matrix}\right)$$
Suppose the point $A$ with position vector $\underline{a}$ is a known fixed point on the plane and the position vector of a general point on the plane is given by $$\underline{r}=\left(\begin{matrix}x\\y\\z\end{matrix}\right)$$
Then for every such point on the plane, we must have $$(\underline{r}-\underline{a})\perp\underline{n}$$
Hence $$(\underline{r}-\underline{a})\cdot\underline{n}=0$$ $$\implies \underline{r}\cdot\underline{n}=\underline{a}\cdot\underline{n}$$ which is a constant $d$, say.
Hence in Cartesian form the equation of the plane is $$ax+by+cz=d$$