how to prove plane $ax+by+cz = d$ has normal vector $(a,b,c)$

Question

Given a plane function $ax+by+cz=d$, how can one prove that unit normal vector is

$$n = \pm \dfrac{a\boldsymbol{i}+b\boldsymbol{j}+c\boldsymbol{k}}{\sqrt{a^2+b^2+c^2}}$$

Answer 1

take two points $A_1= (x_1,y_1, z_1), A_2=(x_2,y_2,z_2).$ then $$ax_1 +by_1 +cz_1 = d, \ ax_2+by_2 + cz_3 = d$$ subtracting one from the other gets you $$ a(x_2-x_1)+b(y_2-y_1) + c(z_2-z_1) = 0$$ which tells you that the $\vec {A_2A_1}$ in the plane $ax+by+cz = d$ is orthogonal to the vector $(a,b,c)^T.$ to make a unit vector, divide by the length $\sqrt{a^2 + b^2 + c^2}.$

Answer 2

Hint: if $p, q$ are two points in the plane, so

$$ ap_1 + bp_2 + cp_3 = aq_1 + bq_2 + cq_3 = d,$$

you want to show that $v = p - q$ is normal to $(a,b,c)$, i.e. that

$$ (a,b,c)\cdot v = a(p_1 - q_1) + b(p_2 - q_2) + c(p_3 - q_3) = 0 $$