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Analytic geometry: equations of planes (checking answers)

  1. Nov 17, 2016 #1
    1. The problem statement, all variables and given/known data
    Let S be a sphere with the equation ##(x-2)^2+y^2+z^2=2 ## and let p a line which satisfies the condition ## p \in (\Pi \cap \Sigma) ## where ##\Pi## and ##\Sigma## are planes with equations:
    ##\Pi :x+z=2##
    ##\Sigma: 5x-2z=3##
    a) Show that S and p have exactly one common point
    b) find the equation of all planes that contain the point ##(0,0,0)##, are parallel to p , do not contain p and touch s at one point

    2. Relevant equations
    plane equation: ## ax+by+cz+d=0##
    line equation : ##\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\theta##
    distance from a point to a plane : ## d=\left | \frac{ax_0+by_0+cz_0+d}{\sqrt{a^2+b^2+c^2}}\right |## where ##x_0,y_0,z_0## are the coordinates of a point and a,b,c are the directional vectors of a plane

    3. The attempt at a solution
    Hi I would really appreciate if some could check my answers of this problem (especially b I think I got that one wrong). Also we just started learning Latex in our uni so I apologize for any mistakes
    so I though a) was pretty easy. What I did was:
    I calculated the directional vector of the line which is equal to ##\vec{p}=\vec{n}_\Pi \times \vec{n}_\Sigma=\begin{bmatrix}
    i&j&k\\
    1&0&1\\
    5&0&-2
    \end{bmatrix}=(0,7,0)=(0,1,0)
    ##
    so now I had the directional vector I just needed a point which I got by solving ##x+z=2## and ##5x-2z=3## and got that x=1 and z=1 so :##p:\frac{x}{0}=1=\frac{z}{0}## and ##y=\theta##
    now I just pluged in x=1, y=##\theta## and z=1 into the sphere equation and got only one common point ##(1,0,1)## can anybody check if this is correct?


    b was kinda harder and I think I got it wrong because the answers I got were not pretty
    b)
    I know that the plane has the point ##(0,0,0)## if we plug this into the plane equation we get ##a(0)+b(0)+c(0)+d=0## which must mean that d =0
    then we also know that since the plane is parallel to the line the scalar product of the lines directional vector and the planes directional vector must be 0 therefore ##(\vec{a},\vec{b},\vec{c})\cdot(0,1,0)=0\Rightarrow b=0##
    so the plane equation know looks like ##ax+cz=0##
    then we also know that the since the plane touches the sphere the shortest distance from the plane to the center is the radius so ## \sqrt{2}=\left | \frac{2a}{\sqrt{a^2+c^2}}\right |## and since ##\sqrt{a^2+c^2}## is the directional vector its length is 1 that means that ##a=\pm\frac{\sqrt{2}}{2}## and since ##\sqrt{a^2+c^2}=1## that means that ##c=\pm\frac{\sqrt{2}}{2}##
    so in the end I got 4 different equations:
    ##\begin{align}
    \frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}z=0\nonumber\\
    -\frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}z=0\nonumber\\
    \frac{\sqrt{2}}{2}x-\frac{\sqrt{2}}{2}z=0\nonumber\\
    -\frac{\sqrt{2}}{2}x-\frac{\sqrt{2}}{2}z=0\nonumber\\
    \end{align}##
    Which is totally weird to me since when I tried to visualize the problem I could only find 2 possible solutions at maximum . So having 4 just seems wrong to me
    thanks for any help
     
    Last edited: Nov 17, 2016
  2. jcsd
  3. Nov 17, 2016 #2

    PeroK

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    Your answer for a) is correct. For b) you went wrong with the scalar products being 0. That's for perpendicular vectors.
     
  4. Nov 17, 2016 #3
    But if a line is parallel to a plane doesn't it mean that is then perpendicular to the normal vector of a plane? or am I missing something
     
  5. Nov 17, 2016 #4

    PeroK

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    Yes, sorry, I missed what you were doing.

    I don't understand what you are doing with the sphere in part b).
     
  6. Nov 17, 2016 #5
    sorry I just noticed I forgot to write something. I forgot to write that the plane should also touch the sphere S at some point .
    thanks for pointing it out
     
  7. Nov 17, 2016 #6

    PeroK

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    You need a diagram. Using geometry rather than algebra might be easier here.
     
  8. Nov 17, 2016 #7
    I'm not sure what you mean. At school we've pretty much only done problems this way .
     
  9. Nov 17, 2016 #8

    PeroK

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    Are you able to visualise this problem at all? Or is it just some equations?
     
  10. Nov 17, 2016 #9
    1 001.jpg
    I hope you can see from the picture, but that is how I envision the problem to look like. And i'm am asked to find ##\tau_1## and ##\tau_2##
     
  11. Nov 17, 2016 #10

    PeroK

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    For example, for part a) you could have recognised that those are vertical planes, drawn the lines in the ##x-z## plane, found the point of intersection and ##p## is the vertical line through that point.
     
  12. Nov 17, 2016 #11

    PeroK

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    I can't see your diagram very well, but it probably doesn't matter. That's geometry! You can solve part b) by using similar geometrical arguments to simplify the algebra required.

    For example, you did a lot of work in part b) to show that a plane parallel to a vertical line is itself vertical!
     
  13. Nov 17, 2016 #12

    PeroK

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    I just looked back at your first post, knowing now about touching the sphere. You got the right answer! Just simplify those four equations and you'll see there are just two really.
     
  14. Nov 18, 2016 #13

    ehild

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    @Matejxx1
    One of the planes contains the line p. The problem asked
    The figure shows the projection of the set-up onto the (x,z) plane.

    upload_2016-11-19_5-50-23.png
     
    Last edited: Nov 18, 2016
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