Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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 $\Pi\cap(\Pi_1\cap\Pi_2)$.)

share|cite|improve this answer
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.