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: Dot & cross products

  1. Oct 26, 2007 #1
    Q:Given non-zero vectors u and v, find a vector x such that both equalitites x dot u=|u|, and x cross u = v hold. What should be the mutual position of the given vectors u and v in space, so that the problem has a solution?

    What I've got so far:

    Setting x = au + b(u cross v)
    I have that
    a=1/|u| and b= +/- |v| / |u cross (u cross v)|

    This answers the first part of the question, but I don't get the second part (i.e. What should be the mutual position of the given vectors u and v in space, so that the problem has a solution?)

    Thanks for helping and explaining!
  2. jcsd
  3. Oct 27, 2007 #2
    This question looks hard and weird, but I am sure someone here knows the answer...
  4. Oct 27, 2007 #3


    User Avatar
    Science Advisor

    I started to say this was impossible until I saw that second question! Just "any" u and v won't do. You know, I expect, that the cross product of two vectors is perpendicular to both. Okay, if x cross u= v, what must be true of u and v?
    Last edited by a moderator: Oct 28, 2007
  5. Oct 27, 2007 #4
    using the second equation,
    [tex] \left( { x \times u } \right) \times u = v \times u [/tex]
    ①[tex] u \left( { x \cdot u } \right) - x \left( { u \cdot u } \right) = v \times u [/tex]
    using the first equation and developping
    ②[tex] x = \frac{1}{\left| u \right|} u - \frac{1}{\left| u \right|^{2}} v \times u [/tex]
    hence x is obtained.
    then considering the following 3 eqations,
    [tex] \left( { x \times u } \right) \cdot u = v \cdot u [/tex]
    [tex] \left( { x \times u } \right) \times v = v \times v [/tex]
    ③[tex] \left( { x \times u } \right) \cdot v = v \cdot v [/tex]
    ④[tex] 0 = v \cdot u [/tex]
    ⑤[tex] u \left( { x \cdot v } \right) - x \left( { v \cdot u } \right) = 0 [/tex]
    l.h.s in ③ [tex] = x \cdot \left( { u \times v } \right) [/tex] so substituting ① into ③
    ⑥[tex] x \cdot \left( { - u \left( { x \cdot u } \right) + x \left( { u \cdot u } \right) } \right) = \left| v \right|^{2} [/tex]
    substituting ④ into ⑤
    [tex] u \left( { x \cdot v } \right) = 0 [/tex]
    ⑦[tex] x \cdot v = 0 [/tex] because [tex] u \neq 0 [/tex]
    according to ④ and ⑦ we can say [tex] u [/tex] and [tex] v [/tex] are orthogonal and [tex] u [/tex] and [tex] x [/tex] are on the same plane perpendicular to [tex] v [/tex] and not parallel due to the given first equation.
    further using ①, ⑥ we can also say the relationship of the norm of [tex] u [/tex], [tex] v [/tex] and [tex] x [/tex] as
    [tex] \left| u \right|^{2} \left( { \left| x \right|^{2} - 1 } \right) = \left| v \right|^{2} [/tex]
    Last edited: Oct 27, 2007
  6. Oct 28, 2007 #5
    I forgot to type the hint that is attached to the orginal question:
    HINT: seek the vector x in the form of a linear combination of u and (u cross v)

    But I'm not sure how this will help...
  7. Oct 28, 2007 #6
    So v and u must be perpendicular, right?
    Is that all?
    Last edited: Oct 28, 2007
  8. Oct 28, 2007 #7


    User Avatar
    Science Advisor

  9. Oct 29, 2007 #8
    But there is another condition that it must satisfy: x dot u=|u|, does this impose any other restrictions?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook