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: Vector projection onto a straight line

  1. Nov 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Determine the matrix for the spatial projection perpendicular to the straight line (x1, x2, x3) = t(1, 2, 3). The vector space is orthonormal.

    2. Relevant equations

    3. The attempt at a solution

    After a trip to #math on freenode that resulted in discussions of Gram-Schmidt processes and bra-ket-like vector notations used in physics, I don't even know where to begin any longer. I suspect that this is supposedly a very rudimentary problem of geometric intuition, since I hadn't heard of projections until yesterday.

    Anyway, if u=(a1,a2,a3) is an arbitrary vector and P is the projection (which is assumed to be linear), then P(u) = a1*P(e1) + a2*P(e2) + a3*P(e3), since e1, e2 and e3 are base vectors. Now, P(e1), P(e2) and P(e3) should be column vectors in the sought matrix.

    Unfortunately, I don't know how to determine their coordinates.

    In one attempt to determine them, I assigned s = (1,2,3) as a vector parallel to the line t(1,2,3). So, the dot product e1*s = |e1||s|cos(A) = 1*1 + 2*0 + 3*0 = 1, if e1=(1,0,0) and A is the angle between e1 and s. Then, A = arccos(1/sqrt(14)), since |s|=sqrt(14).

    However, I'm getting nowhere with this approach. I can't really determine a normal of the line t(1,2,3) either. Bleh. Ideas are welcome.

    Edit: Perhaps I can use the shortest distance between the tip of the vector u and the line...
    Last edited: Nov 19, 2009
  2. jcsd
  3. Nov 19, 2009 #2
    I think I solved it. Not sure if my approach was optimal, but oh well.

    Basically, the sought projected vector should be parallel to s = (1,2,3), e.g. be written on the form L(1, 2, 3), where L is an unknown constant. Then, the vector perpendicular to the line between the tip of u and the line is then b = (L-a1, 2L-a2, 3L-a3). Using the dot product, we get b*s = L-a1 + 4L-2a2 + 9L-3a3 = 0 <=> L = (1/14)*(a1 + 2a2 + 3a3).

    Inserting these values of L gives the coordinates of the sought vector as (1/14)*(a1 + 2a2 + 3a3)*(1,2,3). This corresponds to the projection matrix

    1 & 2 & 3 \\
    2 & 4 & 6 \\
    3 & 6 & 9
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook