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Angle Between Two Lines?

  1. Mar 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the angle between
    a) The line L1 given by the equations y = 2z, x = 0 and
    b) The line L2 given by the equations x = 3z, y = 0.


    2. Relevant equations
    v.u=|v|*|u|*cos(θ)


    3. The attempt at a solution

    I know that I need to basically have a vector for each line to substitute into the equation above, but I am confused as to which set of the following vectors is correct:

    1) L1 can be represented as 0x+1y-2z=0 so the vector/norm of it would be: (0,1,-2) (the tutor in class used norms when someone asked her about the question but I had trouble following what she was doing. :S) L2 would similarly be (1,0,-3). This produces an angle of 31.95°

    2) Let z=1 in both cases, such that: L1 = (0,2,1) and L2 = (3,0,1). This when entered into the formula generates an angle of 81.87°.

    I can see the justification behind both answers but I don't know which would be correct? Please help! :frown:

    Thank you! :smile:
     
  2. jcsd
  3. Mar 10, 2013 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    "Norm" is the wrong word here. For vectors, that refers to the length of the vector which is not relevant.

    No, the line, L1, cannot be represented by "0x+1y-2z=0"- that's the equation of a plane. It is two dimensional because you could choose values for, say, x and y, and solve for the value of z. The vector <0, 1, -2> is perpendicular to that plane.

    As originally given, y= 2z, x= 0 (equivalent to x= 0, y= 2t, z= t) is one-dimensional, a line, because, given a single value of z, we can solve for both x and y. We can write it as a vector equation [itex]\vec{r}(t)= 0\vec{i}+ 0\vec{j}+ 0\vec{k}+ t(0\vec{i}+ 2\vec{j}+ \vec{k})[/itex] showing a little more clearly that the line passes through the point (0, 0, 0) in the direction of the vector [itex]0\vec{i}+ 2\vec{j}+ \vec{k}[/itex] so that vector, or, more simply, <0, 2, 1>, not <0, 2, -1>, points in the direction of the line.

    Similarly, we can write "x= 3z, y= 0" as x= 3t, y= 0, z= t or, in vector form, [itex]\vec{r}(t)= 0\vec{i}+ 0\vec{j}+ 0\vec{k}+ t(3\vec{i}+ 0\vec{j}+ \vec{k})[/itex]. Again, that is a line through (0, 0, 0) but now in the direction of [itex]3\vec{i}+ 0\vec{j}+ \vec{k}[/itex] or just <3, 0, 1>.
     
  4. Mar 10, 2013 #3
    Thanks for the reply, that actually helped quite a bit! :smile:

    Just to confirm - there's no way the lines can be represented as a plane, because that would negate the point of the lines intersecting (if that makes sense).

    Also I tried graphing the equations in MatLab to see which angle seemed more correct and I got something that looked similar to this. After what you said, it seems the method which netted me 81° seems more correct, it appeared in the graphs i made that the 31° answer seems more correct. Have I missed something else, or do I need to get better at graphing 3D lines?

    Thank you!!

    EDIT: This graph shows another view (ignore the axis names, I just switched y and z around so I could see a different view) - so this angle and the other angle should somehow make up 81°? I feel like that's the more correct method... but I'm not entirely convinced since it's matching the evidence I'm seeing in these graphs...
     
    Last edited: Mar 10, 2013
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