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Homework Help: Analytic Geometry Problem - find a line in space meeting some conditions

  1. Feb 3, 2009 #1
    Hi all! I've appreciated (and lurked :-) this forum for a while. While waiting to find a question I'm able to answer, it's time for my first post (hopefully the first of many).

    Here's the problem: I'm trying to teach myself euclidean geometry and a found a tough (for me) question to answer in the textbook I use. As you can read below, I attempted to find a solution but I'm not sure it fits. So I'm asking you help me find out any faults in my approach to the problem. Sorry for the bad english, but I'm just trying to translate:

    Find the equations of the lines lying on the plane [tex]\pi: x-y+z=0[/tex] intersecting the line [tex]r:\left\{\stackrel{x=-z+2;}{y=-2z+2}[/tex] and making a [tex]\frac{\pi}{4}[/tex] angle with the z-axis

    The attempt at a solution
    I tried to solve the problem by noting that the line(s) we are looking for, by lying on the plane [tex]\pi[/tex] and intersecting the line r must share with them one point, given by the intersection of the plane and of the line r itself.

    So I found out it had to pass throug P(2,2,0).

    Then i tought it would be useful to consider the vector [tex]\bar{PZ}[/tex] with Z being the generic point of the z-axis given by (0,0,k).
    The direction of the z-axis is given by [tex]\bar{z}(0,0,1)[/tex].

    So since given two vectors u and v we have that cos uv = [tex]\frac{|uv|}{|u||v|}[/tex], i tried to do [tex]\frac{\sqrt{2}}{2}=\frac{k}{\sqrt{8 + k^{2}}}[/tex]

    Substituting I found out that k=[tex]\pm2\sqrt{2}[/tex]. That's because cos (PZ z) had to be [tex]\frac{\sqrt{2}}{2}[/tex].

    Substituting again, the line(s) I'm looking for should pass from P(2,2,0) and K(0,0,[tex]\pm2\sqrt{2}[/tex]). This gives me two lines, as I had to expect.

    And that's all. I'm just waiting for your suggestions/correction (expecially corrections :-).

    Thanks in advance,

    PS: sorry for bad English and formatting... I have to get used to both :-)
  2. jcsd
  3. Feb 4, 2009 #2
    Hi again...
    I just found a fault in my attempt to solve the problem... I didn't consider that (0,0,[tex]2\sqrt{2}[/tex]) actually doesn lye on the plane [tex]\pi[/tex]!! So i tought to impose that the x and y coordinateof of the point would be such that [tex]x - y + z = 0[/tex], with z being of course [tex]2\sqrt{2}[/tex].
    Given this, everything else seems to be ok.

    Again, thanks in advance for a reply,
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