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Let $\Pi_1 : 7x-5y-2z=0, \Pi_2 : 5x-4y-z=0$ and $\mathbb{L}$ the line that passes through points $P=(-2,3,-3)$ and $Q=(-1,2,-1)$. Find all planes $\Pi$ that satisfy:

  • $\Pi \cap \Pi_1 \cap \Pi_2 = \emptyset$
  • $d(R,\Pi)=\sqrt{14} \forall R \in \mathbb{L}$

So far, I've found out that $\Pi_1\cap\Pi_2= { X : \lambda (1,1,1) } $ and $\mathbb{L} : X=\lambda(1,-1,2)+(-1,2,-1)$; given $\Pi : NX = NQ$, with $N$ being an orthogonal vector to the plane and $Q$ being any point in $\Pi$, it follows that $N \perp(1,1,1)$ and $N\perp(1,-1,2)$, so we can define $N=(1,1,1)\times(1,-1,2)=(3,-1,-2)$.

As $d(R,\Pi) = \frac{|N(R-Q)|}{||N||} = \sqrt{14}$

$\frac{|(3,-1,-2)((-3\lambda-1,\lambda+2,-4\lambda-1)-(x,y,z))|}{||(3,-1,-2)||} = \sqrt{14}$

$|(3,-1,-2)(-3\lambda-1-x,\lambda+2-y,-4\lambda-1-z)| = 1$

$-9\lambda-3-3x -\lambda-2+y +8\lambda+2+2z = \pm 1$

$-3x+y+2z-2\lambda-3 = \pm 1$

and I simply don't know what to do next. Any help would be really appreciated.

EDIT: Solved.

$d(R,\Pi) = \frac{|N(R-Q)|}{||N||} = \sqrt{14}$

$(3,-1,-2)((-1,2,1)-(x,y,z)) = \pm 1$

$-3x+y+2z-7=\pm 1$

So we can take $Q$ as either $(0,4,2)$ or $(0,2,2)$ and we get two different planes:

$\Pi_3: 3x-y-2z=-8$ and $\Pi_4: 3x-y-2z=-6$. Once again, thanks a lot.

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By the way, the wording may be a bit weird since English isn't my first language and the problem was originally written in Spanish. –  F M Sep 2 '10 at 2:37
    
Just to check: d is the plane/line distance? –  J. M. Sep 2 '10 at 4:17
    
$d(R,\Pi)$ is the distance between any point $R$ in line $\mathbb{L}$ and the plane $\Pi$. –  F M Sep 2 '10 at 4:39
    
You might be able to use the fact that the plane $Ax+By+Cz=D$ and the line $(x\;y\;z)=(x_0+a\lambda\;y_0+b\lambda\;z_0+c\lambda)$ are parallel iff $aA+bB+cC=0$. –  J. M. Sep 2 '10 at 5:02
    
In fact, I use that property to set the plane $\Pi$ as parallel to both the line $\mathbb{L}$ and the line formed by $\Pi_1 \cap\Pi_2$. –  F M Sep 2 '10 at 5:23

1 Answer 1

up vote 2 down vote accepted

Martín, I think you can take any value of $\lambda$, because the distance between R and $\Pi$ doesn't change if R is a point of L, since you made sure L is parallel to $\Pi$. I'm not sure, though. (Btw, I suppose the first condition is actually $\Pi\cap(\Pi_1\cap\Pi_2)$.)

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But if $\lambda=k$ for some $k\in\mathbb{R}$, then $Q=(x,y,z)$, that is, any point I like. However, $Q\neq (1,1,1)$ because it would imply that the intersection of the three planes isn't the empty set. I think I screwed up on some of the steps and the lambdas should cancel each other, but I can't figure out where. On my textbook, the problem states $\Pi \cap\Pi_1\cap\Pi_2$ without parentheses, though I think it's the same. Thanks a lot! –  F M Sep 2 '10 at 4:06
1  
Or perhaps this: take any $\lambda$, which corresponds to a point (vector) in L. To this point, add a multiple of the unit vector $\hat{n}$ of magnitude $\sqrt{14}$. You thus get a point particular point of $\Pi$. –  Weltschmerz Sep 2 '10 at 11:34
    
You're totally right, I misunderstood what you said first. Of course, since $\mathbb{L}\parallel\Pi$, if any point in $\mathbb{L}$ is at distance $\sqrt{14}$ from $\Pi$, then all points are. Thanks! –  F M Sep 2 '10 at 18:00

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