To preface the question, with respect to the topic at hand, I am an absolute beginner. I've only very recently dived into a lesson in my book that deals as an intro of sorts to vectors and planes in 3D. Searches in regards to this question on the Internet, if any, only leave me more confused
I've thus far, learned about various ways to represent planes:
- With the normal vector of the plane + distance from the origin : $\vec a \bullet\hat n = d$
- With the normal vector of the plane + a point on the plane : $(\vec r – \vec a) \bullet \hat n = 0$
- With 3 non-collinear points : $(\vec r – \vec a).((\vec b – \vec a) \times (\vec c – \vec a)) = 0$
The fourth one thus far, the intercept form , broaches a new equation for a plane (before moving on to the rest of the derivation) , of which I've not seen before:
$Ax + By + Cz + D = 0$
I'd initially thought that this was a rearrangement of $(\vec r – \vec a) \bullet \hat n = 0$
That is, if $\vec r$ was some vector of the form $x \hat i + y \hat j + z \hat k$, and $\hat n$ was $A \hat i + B \hat j + C\hat k$ , it would yield $Ax + By + Cz – D = 0$, since $\vec a \bullet \hat n$ being the distance from the origin to the plane
However, the equation given in the derivation of the intercept form is
$Ax + By + Cz + D = 0$, wherein we do not find a negative D
How exactly was $Ax + By + Cz + D = 0$ 'derived', so to speak. I can't seem to make sense of it or intuit it
Best Answer
In this second representation, it is not necessary for the normal vector to be a unit normal, so we can write $$(\vec r - \vec a) \cdot \vec n = 0$$ instead. So, we have $$\begin{pmatrix} x-a_1 \cr y-a_2 \cr z-a_3 \end{pmatrix}\cdot \begin{pmatrix} n_1 \cr n_2 \cr n_3.\end{pmatrix}=0,$$ in other words, $$n_1x+n_2y+n_3z-n_1a_1-n_2a_2-n_3a_3=0.$$ Letting \begin{align}A&=n_1\\B&=n_2\\C&=n_3\\D&=-n_1a_1-n_2a_2-n_3a_3,\end{align} the equation becomes $$Ax + By + Cz + D = 0.$$
P.S. The term $$-D$$ could well be a positive number; read the minus sign in front of $D$ not as an adjective/description ("$\require{cancel} \xcancel{\text{negative}} D$") but as a verb/operation ("minus $D$").
Addendum
Read “-5” as “minus 5” instead of “negative 5”. If $D$ equals $(-7),$ then both $(-D)$ and $|D|$ are positive and equal $7.$ Prefixing a number with a minus sign flips its sign; on the other hand, taking the absolute value of a number converts any negative sign to positive.
Your first way of representing planes contains a typo: the equation is $\vec r \cdot\vec{\hat n} = d$ instead or $\vec a \cdot\vec{\hat n} = d.$ Here, $\hat n$ has unit length, and the plane's distance from the origin is $|d|$ instead of $d.$
Similarly, in your second way of representing planes, the plane's distance from the origin is actually $\frac{|D|}{\sqrt{A^2+B^2+C^2}}$ (the denominator equals $1$ if we've been using a unit normal).
To be clear: while the plane's distance from the origin is naturally nonnegative, $d$ and $D$ may be negative or zero or positive.
For example, the planes $x+2y+2z+9=0$ and $x+2y+2z-9=0$ are both $3$ units from the origin.
Notice that "the positive of $(-3)$" can be reasonably interpreted both as +3 and -3; "the negative of $(-3)$" is just as confusing.
I think you mean to ask why $D$ and $d$ have opposite signs. When you rewrite the plane's equation $$Ax + By + Cz + D = 0$$ as $$Ax + By + Cz = D_2$$ (neither equation is more conventional than the other), then $$\dfrac{D_2}{\sqrt{A^2+B^2+C^2}}=d=\dfrac{-D}{\sqrt{A^2+B^2+C^2}}.$$