In the textbook of Differential Geometry by Do Carmo, there is this example:
For a plane $ax+by+cz+d=0$, the unit normal vector is $N=(a,b,c)/\sqrt{a^2+b^2+c^2}$.
I am trying to understand how did they get the unit normal vector.
My try:
I first parametrise the plane by $\textbf{x}(u,v)=(-\frac{b}{a}u-\frac{c}{a}v-\frac{d}{a},u,v)$. Then $\textbf{x}_u=(-\frac{b}{a},1,0)$ and $\textbf{x}_v=(-\frac{c}{a},0,1)$. So $\textbf{x}_u\times \textbf{x}_v=(1,\frac{b}{a},\frac{c}{a})$ and $|\textbf{x}_u\times \textbf{x}_v|=\frac{\sqrt{a^2+b^2+c^2}}{\sqrt{a^2}}$. So $N=\frac{\sqrt{a^2}(1,\frac{b}{a},\frac{c}{a})}{\sqrt{a^2+b^2+c^2}}$.
There is no way I could get $N=(a,b,c)/\sqrt{a^2+b^2+c^2}$ unless $a>0$, but there is no such mention in the problem.
Could somebody please give some light on this? Thanks.
When you get the vector $\vec{x}_u\times\vec{x}_v=(1,b/a,c/a)$ you obtain its length getting $$|\vec{x}_u\times\vec{x}_v|=\sqrt{1+b^2/a^2+c^2/a^2}$$ For now let's say that the length is given by $$|\vec{x}_u\times\vec{x}_v|=\pm\frac{|a|\sqrt{1+b^2/a^2+c^2/a^2}}{a}$$ $$=\pm\frac{\sqrt{a^2}\sqrt{1+b^2/a^2+c^2/a^2}}{a}$$ $$=\pm\frac{\sqrt{a^2+b^2+c^2}}{a}$$ (We can do this because $|a|/a=\pm 1$ depending on the sign of $a$). Then dividing our vector $\vec{x}_u\times\vec{x}_v$ by its length to normalize it we get $$\vec{N}=\frac{\vec{x}_u\times\vec{x}_v}{|\vec{x}_u\times\vec{x}_v|}$$ $$=\pm\frac{a(1,b/a,c/a)}{\sqrt{a^2+b^2+c^2}}$$ $$=\pm\frac{(a,b,c)}{\sqrt{a^2+b^2+c^2}}$$ Which as you can see means the issue at hand just boils down to the choosing the orientation of the normal vector, which by convention we choose it by picking the plus sign in the above equation.