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Homework Help: Determine the point on the Line r = p+tv closest to the origin

  1. Aug 27, 2011 #1
    Find an expression for the distance between the origin and the line L given by r = p + tv, where t is an element of R and determine the point on L closest to the origin

    So the first part of the question i took as just the magnitude of p since the distance will just be |OP|=|P|=(Xo^2+Yo^2+Zo^2)^1/2.

    However finding the point on L closest to the origin I'm finding very confusing, I've tried many things such as minimizing |P|=(Xo^2+Yo^2+Zo^2)^1/2 which was too hard for me.

    Also the dot product of PR & R = 0, where P is a point on the Line and R is the point on the line closest to the origin. Which gave ax+by+cz=0, where a,b,c are components of the direction vector and x,y,z are the point closest to the origin. This is a solution but it's not unique so that can't be right.

    Just a few things I've tried...


    Please help!


  2. jcsd
  3. Aug 27, 2011 #2
    Do you see that the shortest distance between the line and the origin will be given by the normal of the line through the origin?

    So you just need to find a normal vector of the line. Then you take the line through the origin and this normal vector and you see where it intersects this line...
  4. Aug 27, 2011 #3
    Thanks for replying,

    Yea the shortest distance will be a normal to the line that passes through the origin, so will that will be a normal to the direction vector of the line,r, that also passes through the origin. But how do I find this?


  5. Aug 27, 2011 #4
    So you need to find a vector n that is normal to v. That is, you must find n such that

    [tex]n\cdot v =0[/tex]

    where [itex]\cdot[/itex] denotes the inproduct. Can you find such an n?
  6. Aug 27, 2011 #5
    This will give
    ax+by+cz=0, where x,y,z are components of the normal vector and a,b,c components of the direction vector of the line.

    but this is not unique?
  7. Aug 27, 2011 #6
    No, this is not unique. You will have an entire plane of normal vectors!! This plane is uniquely described by ax+by+cz=0. Now, what is the intersection of this plane and the original line?
  8. Aug 27, 2011 #7
    Would this be ax+by+cz=p+tv?
    Can I break this into components?

    Is it okay to assume that point P is the point we are interested in?
  9. Aug 27, 2011 #8
    I know what you mean and it is right. But you can't do what you do here. p+tv is a vectorial equation, that is: p and v are vectors. But ax+by+cz are not vectors. So you can't just set them equal.

    What you do have is that you're vectorial equation p+tv defines a system

    [tex]\left\{\begin{array}{c} x=x_0+ta\\ y=y_0+tb\\ z=z_0+tc\end{array}\right.[/tex]

    Now plug in these x, y, z into ax+by+cz=0 and solve for t.
  10. Aug 27, 2011 #9
    Thanks for helping so much!

    t=-(v.p)/(a^2+b^2+c^2) - don't know if this is necessary?

    So are the points Xo,Yo,Zo now the points we are interested in?


  11. Aug 27, 2011 #10
    You're not there yet. You have found t. Now you got to plug that into the equation of your line to find the intersection. This will be your answer.
  12. Aug 27, 2011 #11
    So this is going to be very general but:

    r = p+tv
    (x,y,z) = (Xo,Yo,Zo)-(aXo+bYo+cZo)/(a^2+b^2+c^2)(a,b,c)


    Now is that it? hah

    Thanks again.
  13. Aug 27, 2011 #12
    That seems to be it, yes :smile:

    A little shorter (but equivalent) expression is

    [tex]p+\frac{v\cdot p}{\|v\|^2}v[/tex]
  14. Aug 27, 2011 #13
    Oh wow that's much more simplified. Thanks so much! :)
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