1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Distance of line to origin

  1. May 9, 2013 #1
    1. The problem statement, all variables and given/known data



    Say you have an equation of a line in 3 space and you want to find the distance from the line to the origin how do you go about it?
    I guess by definition the distance of a line to a point ( say the origin in this case) is the minimum distance such that the line and some vector are perpendicular. That kind of sucks how I said it.

    So x is a unique point on the line so you want the distance xo ( where o is the origin) such that xo is perpendicular to the line L . Not really sure how to do this. It is in the cross product section so maybe that would work. I tried to use dot product but I don't know how to get the shortest point. If I did I should dot xo with L and it would be 0. Well that was my thinking. I don't really know how to go about this.
    Thanks
     
  2. jcsd
  3. May 9, 2013 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In general: Draw a line L and label a point P on the line. Draw a direction vector D for the line with its tail at P. Take a point Q off the line and draw a vector V=PQ. Drop a perpendicular from Q to the line and label its length h. Label the angle between V and D as ##\theta##. What is the length of h using that little triangle you see. The formula you get should remind you of something about the cross product. See if that helps when you apply it to your problem.
     
  4. May 9, 2013 #3
    I don't understand. How does it guarantee it is the shortest distance that is perpendicular and goes through the origin?
    I don''t understand this
    Take a point Q off the line and draw a vector V=PQ. Drop a perpendicular from Q to the line and label its length h.
    Thanks
     
  5. May 9, 2013 #4

    Mark44

    Staff: Mentor

    Edited to include the quote from LCKurtz. Click the Quote button to see how I did it.
     
  6. May 9, 2013 #5

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If you draw any other line K from the point Q to your given line L, then K, L, and the line labeled h make a right triangle with K as the hypotenuse and h a leg. h will always be less than the length of K. Draw a picture.
     
  7. May 9, 2013 #6
    I apologize I can't follow you. I grew a picture but I can't connect it to my problem I don't see how an equation for h is related to the cross product.I would think it is Pythagoras to find h. But once again to me it seems to far away from my problem I can't draw this together.
     
  8. May 9, 2013 #7

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Can't you use a triangle trigonometry formula to write h in terms the length of V and the variable ##\theta##?
     
  9. May 9, 2013 #8
    Vsinθ = h
    PQsinθ = h

    OK but at the same time this line is not so pretty as the one you have described.
    My position vectors are a = <2 , 1, 7> : b = <1 , 4, -1>
    Just connect points A and B this is the line. Thanks
     
  10. May 9, 2013 #9

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I'm not worried about your particular values at the moment. I'm trying to show you what your problem has to do with a cross product. You have ##|V|\sin\theta = h##. Notice that is the length of ##V##, not ##V## itself in the formula. Now in your picture, imagine D being divided by its length giving a unit vector ##\hat D##. Draw a little short vector on top of ##D## representing ##\hat D##. What is the formula for ##|V \times \hat D|##, the magnitude of the cross product of those two vectors?
     
  11. May 9, 2013 #10
    How do you write out your answers really good like that ? Like how the symbols are in your last post?
    VDsin(theta) is the magnitude of the cross product of the v cross d.
    So what does Vsin(theta) have to do with it?
    OK I drew it. Still not landing this concept. I don't get this problem. Thanks
     
  12. May 9, 2013 #11

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Remember ##\hat D## is a unit vector so you have$$
    |\vec V \times \hat D|=|\vec V||\hat D|\sin \theta|$$Now since ##\hat D## is a unit vector, its length is ##1##. So you have$$
    |\vec V \times \hat D|=|\vec V||\hat D|\sin \theta=|\vec V| \sin\theta = h$$Now how does this help you with your problem? Using your given position vectors, you can make a direction vector ##\vec D## and make a unit vector ##\hat D## out of it. What can you use for P and Q in your problem so you can calculate ##\vec V = \overline{PQ}##? Once you have ##\hat D## and ##\vec V##, just calculate ##\vec V \times \hat D## and take its length to get h.

    To read about typing equations, read the pinned FAQ at the top of this forum.
     
  13. May 9, 2013 #12
    Can I just use position vector a as PQ?
     
  14. May 9, 2013 #13

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, because the point that you are interested in that is not on the line is the origin. That is your Q. It doesn't matter if you use PQ or QP because only the length of the vector matters in the formulas.
     
  15. May 15, 2013 #14
    I feel like this problem isn't correct. The approach I mean.
     
  16. May 15, 2013 #15

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In post #8 you seemed to understand from the triangle ##|\vec V|\sin\theta = h##

    And we have$$
    |\vec V \times \hat D|=|\vec V||\hat D|\sin \theta=|\vec V|\cdot1\cdot \sin\theta $$so it gives ##h##. It doesn't matter what you "feel like".
     
  17. May 15, 2013 #16
    Yeah I understand that. Is there a way to draw on here? How can you show the diagram.
    I think if you cross the position vectors it give you a vector orthogonal to the plane. Then you can cross that with the vector that is give by subtraction the position vectors ( what you call D). That will form a new vector in the plane of a and b. You turn it into a unit vector and dot it with the position vector b.
     
  18. May 15, 2013 #17

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    And what does dotting it with the position vector b get you? Nothing relevant that I can see.

    You are making it much harder than it is. You have two points ##a## and ##b## determining the line and a point ##p## not on the line. You can think of all three as position vectors to the points or coordinates of the points.$$
    \vec D = b-a$$ $$
    \vec V = p-a$$Both of those are trivial to calculate. Make a unit vector ##\hat D## out of ##\vec D## by dividing it by its length. Now calculate ##\vec V \times \hat D## and calculate its length and you are done.
     
  19. May 15, 2013 #18
    Dotting with b will get you the length. You say that a is perpendicular to the line?
     
  20. May 15, 2013 #19

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No it won't and no I didn't say that and no it isn't.
     
  21. May 15, 2013 #20
    I think you are finding a different length.

    Forgive my word drawing but it works. I was looking for the blue length. I think it will show up blue.

    So,

    a x b = V , this is a vector orthogonal to the plane.
    AB x V = U
    This vector is in orthogonal to AB.
    So we make a unit vector out of it. Call it u.
    Then we take the dot product of u with b which will give us the length.
    Could you open the drawing?
     

    Attached Files:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted