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!

Homework Help: A question about vectors

  1. Apr 10, 2007 #1
    1. The problem statement, all variables and given/known data

    1) Find the equation of the line L which passes through the point P0(1, 0, 2) and
    intersects the line L1 :(x − 1)/2=(y + 1)/3, z = 1 orthogonally.

    3. The attempt at a solution

    ı thought the line L1 as a parametrisation like x=2t+1;y=3t-1;z=1; then I took a vector parallel to L as ai+bj+ck. Then since the parallel vector of L1 (which is 2i+3j) orthogonal to ai+bj+ck then the dot product of them will 0. This lead 2a+3b=0; Therefore, I can think L as ; x=1+3bt/2 and y=bt. However, my question is how about c? how can ı relate it to a and b? Or is my approach to question is wrong?
  2. jcsd
  3. Apr 10, 2007 #2


    User Avatar
    Science Advisor

    That's perfectly valid but there is no way to relate c to a and b. First, notice that t is not "unique". Multiplying t by any number just changes the length of the parallel vector giving different parametric equations for the same line.

    The line x= 1+ 3bt/2, y= bt, z= ct+ 1 is orthogonal to L1 for all b and c but does not necessarily intersect it. In order to intersect it, for some t, you must have (x-1)/2= 3bt/4= (y+1)/2= (bt+1)/3, z= ct+1= 1. From the last, either c= 0 or t= 0. If you put t= 0 into 3bt/4= (bt+1)/3 you get 1/3= 0 which is not true. Therefore, c= 0 and z= 1 for all t. Now, since t is arbitrary, assume your parmeterization is such that your new line intersects L1 when t= 1. Solve 3b/4= (b+1)/3 for b.
  4. Apr 10, 2007 #3
    it should z=ct+2 isn't it? Because the vector passes the point (1,0,2). And if it is so then ct+2=1 and ct=-1; then we could not fnd exact values to c or t . then how could we continue? ;
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook