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!

Find unknown vector X if these relations hold true

  1. Jan 29, 2015 #1
    • Member warned about posting a problem with no effort shown
    1. The problem statement, all variables and given/known data
    If an unknown vector X satisfies the relation
    X · b = β
    X × b = c
    express X in terms of β, b, and c.


    2. Relevant equations
    X · b = |X||b|cos(θ)
    X × b = |X||b|sin(θ)
    3. The attempt at a solution
    I don't know where to start... :( someone pls give me a hint
     
  2. jcsd
  3. Jan 29, 2015 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    For the cross product, you need notation to say "the magnitude of " [itex] | X \times b | = ....[/itex]

    One approach to this problem would be to represent the vectors in their 3 components (e.g. [itex] (X_x,X_y,X_z) [/itex]) but I think that would be a big mess.

    Another approach is to assume that "express [itex] X [/itex] in terms of [itex] \beta,b,c [/itex] can be fulfilled by a treating those quantities as single symbols. With that approach we can write [itex] X = P + Q [/itex] with [itex] P, Q [/itex] vectors as long as they are expressed in terms of the symbols [itex] \beta, b, c [/itex]. Begin by letting [itex] P [/itex] be the component of [itex] X [/itex] that in in the same direction as vector [itex] b [/itex]. The direction of [itex] b [/itex] is expressed as the unit vector [itex] \frac{b}{|b|} [/itex]. Can you find the component of [itex] X [/itex] in that direction? Perhaps you have studied how to "project" a vector onto another vector.

     
  4. Jan 30, 2015 #3
    OK that makes sense! so I'm guessing Q would be the projection of X onto c? However I'm stuck from here because i cannot express that projection in terms of correct variables (don't know A dot C)..... am i wrong?
     
  5. Jan 30, 2015 #4

    Stephen Tashi

    User Avatar
    Science Advisor

    It wouldn't be the projection on c because c is perpendicular to the plane of X and b. To get something in the same plane as X and b, try the vector (b x c).
     
  6. Jan 30, 2015 #5
    isn't (b x c) = X?
     
  7. Jan 30, 2015 #6

    Stephen Tashi

    User Avatar
    Science Advisor

    Suppose b = the unit x vector and X = (1,1,0). Then X x b = c has the direction of (-1) times the unit z vector. And b x c has the direction of the unit y vector.
     
  8. Jan 30, 2015 #7
    OK that makes sense.... but how would i express X dot (B x C) when i am formulating the projection? I'm trying to take X out... :/
     
  9. Jan 30, 2015 #8

    Stephen Tashi

    User Avatar
    Science Advisor

    My guess is that you'll have to use one of the formulas that expresses X dot (b x c) in a different way. ( perhaps the signed volume of a parallelopiped formula in http://en.wikipedia.org/wiki/Cross_product ) See if there is an expression that uses X x b or b x X since you can replace those expressions with c or -c. Don't forget that projecting X on a vector V involves dividing (X dot V) by |V|.
     
  10. Jan 30, 2015 #9
    Thanks so much. i finally got the answer... I appreciate you tremendously!!!!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted