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Homework Help: Unit vector perpendicular to two known vectors

  1. Jan 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Find a unit vector which is perpendicular to both of the vectors a = 4i + 2j - 3k and b = 2i - 3j + k

    c = xi + yj + zk


    2. Relevant equations
    a[tex]\bot[/tex]c [tex]\longrightarrow[/tex] a [tex]\bullet[/tex] c

    3. The attempt at a solution
    Okay, here's what I've done so far.

    Take the dot-product of a and c, and b and c
    a [tex]\bullet[/tex]b: 4x + 2y -3z = 0
    b [tex]\bullet[/tex]b: 2x - 3y + z = 0

    (1) 4x + 2y -3z = 0
    (2) 2x - 3y + z = 0

    I isolate z and get rid of x by multiplying (1) with -2 and (2) with 4, then add them:
    (1) -8x -4y = -6z
    +
    (2) 8x -12y = 5z

    -16y = 10z

    y/z = 10/16

    Which again means that:
    y = 10m
    z = 16m
    where m is a constant and [tex]\neq[/tex] 0

    And then I insert this into (1) to find x:
    4x + 2(10m) - 3(16m) = 0
    4x + 20m - 48m = 0
    x = 7m

    c = m(7i + 10j + 16k)

    For the easiest possible solution, m = 1.
    c = 7i + 10j + 16k
    As far as I can tell, this is a perfectly valid answer.

    However, the answer key has the answer:
    (1/9[tex]\sqrt{5}[/tex])(7i + 10j + 16k)

    While this does not contradict my solution, that is a far too weird m to have been chosen randomly. Does anyone see how they were thinking?

    Thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 10, 2010 #2
    The question asked for a unit vector ie a vector with magnitude of 1. Hence m must be chosen so that the magnitude of the vector is one.
    Also, there is a much much easier way of solving the problem. The cross product of two vectors yields a third vector that is perpendicular to both original vectors. In this case, simply take the cross product with a with b, and adjust to obtain a unit vector.
     
  4. Jan 10, 2010 #3
    *facepalm*

    Yes, of course.

    And this exercise is from the sub-chapter before cross products, so I figured I'd try to do it the way they wanted me to do it.

    Thanks a lot :)


    Edit: Just to complete the solution:

    | c | = [tex]\sqrt{7^{2} + 10^{2} + 16^{2}}[/tex] = [tex]\sqrt{405}[/tex]
    = [tex]\sqrt{81}[/tex][tex]\sqrt{5}[/tex] = 9[tex]\sqrt{5}[/tex]

    Thus m = 1/ 9[tex]\sqrt{5}[/tex]
     
    Last edited: Jan 10, 2010
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