Finding the vector perpendicular to the plane

Solution 1:

Take two points on the plane: $(x_1,y_1,z_1),(x_2,y_2,z_2)$. Then they both satisfy the plane equation: $$2x_1-y_1+3z_1=8,$$ $$2x_2-y_2+3z_2=8.$$This gives $\left < x_1-x_2,y_1-y_2,z_1-z_2 \right > \cdot \left < 2,-1,3 \right > =0$. In other words, any vector on the plane is perpendicular to the vector $\left < 2,-1,3 \right >$.

Solution 2:

I am assuming we are on $\mathbb{R}^{3}$.

A plane is determined by a point on the plane and a vector orthogonal to the plane. Say $P_0$ is a point on this plane, and $\vec{n}$ is the orthogonal vector. Also let's show the position of point $P_0$ with the vector $\vec{r_{0}}$.

Now take a generic point on the plane, call it $P$, show its position with the vector $\vec{r}$. Then the following equation has to be satisfied:

$\vec{n}.(\vec{r} - \vec{r_{0}}) = 0$

This has to hold because the difference vector $\vec{r} - \vec{r_{0}}$ has to lie in our plane.

Now let's try to find out the scalar equation for our plane. First start by substituting our vectors:

$\vec{n} = \langle a, b, c \rangle$, $\vec{r} = \langle x, y,z \rangle$ and $\vec{r_{0}} = \langle x_{0}, y_0, z_0 \rangle$.

Expanding our vector multiplication $\vec{n}.(\vec{r} - \vec{r_{0}}) = 0$ will give us:

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

If you define $d = x_0 + y_0 + z_0$, then the previous equation becomes:

$ax + by + cz = d$.

From this equation you can identify our normal vector $\vec{n}$ directly from the coefficients $\langle a, b, c \rangle $ of the scalar equation of plane.