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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.

    • 0
      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)2010-12-02
    • 0
      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.2010-12-02
    • 2
      Welcome! and +1 for showing the work you did.2010-12-02
    • 0
      @Moron thank you for the welcome.2010-12-02
    • 0
      @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.2010-12-02

    3 Answers 3