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this is me again, I have another problem which I haven't been able to solve, the legend goes like this:
Find the equation of the plane that contains the points $P_1(1,0,-1)$, $P_2(2,0,2)$ and forms a 600 ($\theta$) angle with the other plane $2x-2y+z+6=0$ (Two Solutions).

Here's what I've done:

Since we already know two points, P1(1,0,-1) and P2(2,0,2) they both satisfy a planes equation:
$$Ax+By+Cz+D=0$$ So, we substitute $x,y,z$ respectively for each point, giving use these two equations

  1. $A(1)+B(0)+C(-1)+D=0$
  2. $A(2)+B(0)+C(2)+D=0$

Simplifying we get:

  1. $A-C+D=0$
  2. $2A+2C+D=0$

Now, according to the following theorem the angle that is formed by the normals of to planes is the following:

$$\cos{\theta} = \pm \frac{AA'+BB'+CC'}{\sqrt{A^2+B^2+C^2} \sqrt{(A')^2+(B')^2+(C')^2}}$$ Where $[A,B,C]$ Are the director numbers of the normal in the first plane and $[A',B',C']$ Are the director numbers of the normal in the second plane.

Now, given those and taking the known angle of 60o, we substitute the director numbers of $2x-2y+z+6-0:[A'=2,B'=-2,C'=1]$ and the angle in the last formula, giving us: $$\cos{60^\circ} = \pm \frac{A(2)-B(-2)+C(1)}{\sqrt{A^2+B^2+C^2} \sqrt{(2)^2+(-2)^2+(1)^2}}$$

Simplifying, we get(positive because $\theta$ is acute): $$\frac{1}{2}=\frac{2A-2B+C}{3 \sqrt{A^2+B^2+C^2}}$$

Another simplification finally for: $$3(A^2+B^2+C^2)=(4A-4B+2C)^2$$ Now, we have 3 equations:

  1. $A-C+D=0$
  2. $2A+2C+D=0$
  3. $3(A^2+B^2+C^2)=(4A-4B+2C)^2$

If we take 1) and 2) and write them as $D(A,C)$ we get:

  1. $D=C-A$
  2. $D=-2A-2C$

From which we can obtain A in terms of C $$-2A-2C=C-A$$ $$A=-3C$$ Now we substitute back in 1): $$(-3C)-C+D=0$$ $$C=\frac{1}{4}D$$ Since I want every director number in terms of D we substitute again but in A since A was in function of C. $$A=-3C$$ $$A=-3(\frac{1}{4}D)$$ $$A=\frac{-3}{4}D$$

Now, we have found $A(D)$, $C(D)$ we have yet to find $B(D)$ I'll explain why I think and want everything in terms of D, if we take the general form of a planes equation $Ax+By+Cz+D=0$ and substitute $A$ and $C$ with our known values we get: $$(\frac{-3}{4}D)x+By+(\frac{1}{4}D)z+D=0$$ if we divide by $D \neq 0$ and then multiply by 4 to eliminate fractions: $$\frac{-3}{4}x+\frac{B}{D}y+\frac{1}{4}z+1=0$$ $$-3x+\frac{4B}{D}y+z+4=0$$ From here it's clear that if we find B in terms of D and B is linear we have the planes equation, so, in 3) $3(A^2+B^2+C^2)=(4A-4B+2C)^2$ we substitute A and C: $$3((\frac{-3}{4}D)^2+B^2+(\frac{1}{4}D)^2)=(4(\frac{-3}{4})-4B+2(\frac{1}{4}D))^2$$ $$\vdots$$ $$104B^2+80DB+35D^2=0$$

Here is were I get stuck, I get imaginary parts, since I have to use the quadratic formula, so any help is really appreciated, the problem states that there are two solutions, but this is what I have thinked of if anyone can supply any solution I'll be very gratefull.

Regards.. Tristian

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+1 for showing what you've done. –  J. M. Dec 5 '10 at 9:50
1  
@J.M. Hehe thanks I find it fun to use Tex (or LaTeX) syntax, love to see it render.. –  Triztian Dec 5 '10 at 9:57

2 Answers 2

The normal vector to the sought plane is orthogonal to the line $P_1 P_2$ and at angle $\pi/3$ to the normal to your given plane.

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Well, your statement helped me find $A(C)$ using the perpendicularity theorem, but it just gives me a known fact: Director numbers of $l:[1,0,3]$ $$A(1)+B(0)+C(3)=0$$ we get: $$A=-3C$$ –  Triztian Dec 5 '10 at 10:17
    
That follows from orthogonality; now use $60^\circ$. –  Robin Chapman Dec 5 '10 at 10:22
    
That's where I lost you, how can it be orthogonal at $60^\circ$, if in order to be orthogonal it must be a $90^\circ$, where should I use the $60^\circ$?, thank you for your reply and patience by the way. –  Triztian Dec 5 '10 at 10:39
    
@Triztian: It's orthogonal to one vector and at 60 degrees to another vector. There's no contradiction in that. Just to exemplify: the vector (cos 60, sin 60, 0) is at 60 degrees to (1,0,0), and at the same time orthogonal to (0,0,1). –  Hans Lundmark Dec 5 '10 at 18:48
    
@Hans Lundmark, thanks for the explanation,now I get it, but after reading and reading Robin Chapman's statement, it seems that he is saying something that the problem says in it's legend, that is: "That the director numbers of the plane that I'm looking for are orthogonal to the line segment $P_1 P_2$ and that it forms a $60^\circ$ (\frac{\pi}{3}) angle with the known plane's normal... or Maybe I'm missing the point, did anyone look at what I already did? –  Triztian Dec 7 '10 at 6:56

To flesh out Robin's answer a little: The thing that you really want to find is the unit normal vector $n=(A,B,C)^t$ to the sought plane. Once you have that, it's easy to find the equation of the plane. This vector has to satisfy three conditions:

  1. Length one: $A^2+B^2+C^2=1$.

  2. Orthogonal to the vector $\overrightarrow{P_1 P_2} = (1,0,3)^t$.

  3. 60 degrees to the normal vector of the other plane. If we normalize that vector too, this condition becomes $\frac13 (2,-2,1)^t \cdot (A,B,C) = \cos 60^{\circ}$.

Conditions 2 and 3 give a linear system with two equations and three unknowns $A$, $B$, $C$; solve it and you will get a family of solutions depending on one parameter $t$. Condition 1 then gives a quadratic eqation for determining the two possible values of $t$.

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