A plane $P$ contained in $\mathbb R^3$ is given by the equation $ax +by +cz+d=0$. Show that the vector $v = (a, b, c)$ is perpendicular to the plane.

Question

This problem is from do Carmo. In calculus, we are given this vector orthogonal to the plane and derive the equation $ax+by+cz+d=0$, but here the question seems to be asking the opposite.

I'm not sure how to go about the question. Any hints?

Answer 1

Let $P_1 = \left(x_1, y_1, z_1\right)$ and $P_2 = \left(x_2, y_2, z_2\right)$ be any $2$ distinct points in the plane. Then $P_1 - P_2 = \left(x_1 - x_2, y_1 - y_2, z_1 - z_2\right)$ is a vector in this plane. Taking the dot product of this with $v = \left(a, b, c\right)$ gives

$$a\left(x_1 - x_2\right) + b\left(y_1 - y_2\right) + c\left(z_1 - z_2\right) \tag{1}\label{eq1}$$

As $P_1$ and $P_2$ are on the plane, they must satisfy the equation $ax + by + cz + d = 0$. Thus, this gives

$$ax_1 + by_1 + cz_1 + d = 0 \tag{2}\label{eq2}$$

and

$$ax_2 + by_2 + cz_2 + d = 0 \tag{3}\label{eq3}$$

Subtracting \eqref{eq3} from \eqref{eq2} gives

$$a\left(x_1 - x_2\right) + b\left(y_1 - y_2\right) + c\left(z_1 - z_2\right) = 0 \tag{4}\label{eq4}$$

Thus, since the dot product is $0$, then the $2$ vectors are perpendicular. As $P_1$ and $P_2$ are any $2$ points, this shows the vector $v$ is perpendicular to the entire plane.

Answer 2

Pick any vector $w = (m,n,p)$ on the plane $P.$ Then, you need to prove that: $$(v,w) = am+bn+pc = 0.$$ Now, a vector on a plane is really the same as two points on a plane, call them $ Q_i = (x_i, y_i,z_i)$ with $i=1,2.$ Then, what is $m,n,p$ in terms of these coordinates? Don't also forget that $Q_i$ satisfies the equation of the plane.

Can you follow it from here?