I'm reading Apostol's Calculus.
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And I've tried to do the following exercise:
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I am a little confused: I have the portuguese version of the book, and it says $P=F+dN$, in this english version, it says $dP=F+dN$. I've tried to think with my version of the book. I did the following:
$$(p_1,p_2)=(f_1,f_2)+d(n_1,n_2)\\(p_1,p_2)-(f_1,f_2)=d(n_1,n_2)\\(p_1,p_2)-(f_1,f_2)=(dn_1,dn_2)\\dn_1=p_1-f_1 \quad \quad \quad \quad \quad \quad \quad dn_2=p_2-f_2 \\ \text{As $N$ is the normal unit vector, I guess I can assume that:}\\d=\frac{p_1-f_1}{n_1} \quad \quad \quad \quad \quad \quad \quad 0=p_2-f_2 $$
EDIT: There is one more thing I got from the book, that $|d|=\|P-F \|$, then I guess that $d=\pm \|P-F \|$, and also that $P=F+dN$, then I can write:
$$\|X-F \|=e|(X-F)\cdot N-d|$$
As:
$$\|X-F \|=e|(X-F)\cdot N-\|\overbrace{F+dN}^{P}-F \||$$
$$\|X-F \|=e|(X-F)\cdot N-\|dN \||$$
I guess that from this, it follows that $d$ must be negative if $F$ is in the left side of the directrix and must be positive is $F$ is on the right side of the directrix. Mainly because $F=P-dN$, if $F$ were on the other side of the directrix, it would need to be $F=P+dN$.