# Analytic Geometry Doubt

Hello all this is my first math question, I hope this is the right place to post. Anyway, this is basic stuff but I'm a bit confused here, here's the problem's legend: "Find the equation of the plane that passes through the intersection line ($l$) of the planes $2x-y-z=2, x+y-3z+4=0$ and has a distance of 2 from the origin.

Ok, So here's what I've done, the family of planes that pases through $l$ is $$2x-y-z-2+k(x+y-3z+4)=0$$ We work the equation so it looks more like a planes general form: $$Ax+By+Cz+D=0$$ giving us this:$$(2+k)x+(k-1)y+(-3k-1)z +2(2k-1)=$$ where$$A=(2+k)$$ $$B=(k-1)$$ $$C=(-1-3k)$$ $$D=2(2k-1)$$

The problem states that it's distance to the origin is 2 and the normal form of a plane($x\cos{\alpha}+y\cos{\beta}+z\cos{\gamma}-p=0$) where $p$ is the distance of the plane to the origin. We know that to transform a planes equation from the general form to the normal form we must divide each term by $$r=\pm \sqrt{A^2+B^2+C^2}$$ So $p=\frac{D}{r}$ and we have this:$$2=\frac{2(2k-1)}{\pm \sqrt{(2+k)^2+(k-1)^2+(-1-3k)^2}}$$ And here is where my doubt is, what sign should $r$ have positive or negative, the following theorem says:

• if $D \neq 0$, $r$ is opposite sign of $D$
• if $D=0$ and $C \neq 0$ $r$ is equal sign of $C$
• if $D=0$ and $C=0$ and $B \neq 0$ $r$ is equal sign of $B$
• if $D=0$ and $C=0$ and $B=0$ and $A \neq 0$ $r$ is equal sign of $A$
• Since we don't know the value of k and $D(k)=2(2k-1)$ and to have $D$ we must find k but k is determined by the sign of D(what a mess hehe) how do we chose $r$'s sign?

Any help is really appreciated.

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I don't see how you got your values of $A$, $B$, $C$, and $D$. I get $(1+k)x + (k-1)y +(-3-k)z + (4k-2) = 0$, or $A=1+k$, $B=k-1$, $C=-3-k$, and $D=4k-2$. Even taking your values as true, your $B$ and $C$ are incorrect (should be the negative of what you have) –  Arturo Magidin Dec 2 '10 at 6:49
Okay, that explains $A$; you still your equation wrong, though; the equation has the sign of $B$ wrong, and your expression for $C$ has the wrong sign. –  Arturo Magidin Dec 2 '10 at 7:03
Welcome! and +1 for showing the work you did. –  Aryabhata Dec 2 '10 at 7:04
@Moron thank you for the welcome. –  Triztian Dec 2 '10 at 7:10
@Arturo Magidin Yeah, corrected again the equation, sorry, it was a bit awkward writing in Tex, I hope I don't have any errors now, thank you for your patience. –  Triztian Dec 2 '10 at 7:12

First of all, your equation should be $$(k+1)x+(k-1)y-\cdots.$$

Note that the distance formula is $$2=\frac{|2(2k-1)|}{\cdots}$$ (no $\pm$). Square both sides (note that square of $|\text{anything}|$ is $\text{anything}^2$, then solve a quadratic equation. You will get two $k$'s. Substitute back to see which one is correct. In this case, it turns out that both $k$'s are correct.

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why no $\pm$ sign, the process of transforming from the general form to the normal form involves dividing by $r=\pm \sqrt{A^2+B^2+C^2}$, thats how we know that $$2=\frac{2(2k-1)}{\pm \sqrt{A^2+B^2+C^2}}$$ and the sign of $r$ is determined by the theorem I mention. –  Triztian Dec 2 '10 at 7:21
$$2=\frac{|2(2k-1)|}{\sqrt{A^2+B^2+C^2}}$$ is better than $$2=\frac{2(2k-1)}{\pm \sqrt{A^2+B^2+C^2}}$$ since the latter one is misleading; distance cannot be negative. –  TCL Dec 2 '10 at 14:11
You are completely right, but in this case the distance of the single line to the origin doesn't matter because the line is just used a as reference to find the equation of the plane that we're looking for, if I had the line's distance to the origin I still wouldn't be able to calculate the director numbers $[A,B,C]$ of the plane. –  Triztian Dec 2 '10 at 6:36