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Algebraic vector problems (planes, cross products etc)

  1. Nov 1, 2012 #1
    1. The problem statement, all variables and given/known data

    1. A vector u is given by u = λa + μb, where λ, μ are elements of ℝ. Calculate axu and bxu and show that:

    [itex][(b χ u).n]/[(b χ a).n][/itex] = λ

    and

    [itex][(a χ u).n]/[(a χ b).n][/itex] = μ

    (In any circumstance when I have omitted the underlining it is for convenience purposes, a, b, n and u are still vectors.)


    2. The vectors a and b form a set of basis vectors for P1 (A plane spanned by vectors a and b). However it is often desirable to use a set of orthogonal basis vectors.

    Construct a vector b' that is in P 1 and orthogonal to a.

    2. Relevant equations

    We are told that a and b are vectors in ℝ3, and that they are not parallel or anti-parallel. n is the normal vector to plane P1.

    3. The attempt at a solution

    1. n= a x b

    a x b = a x (aλ x bμ)
    = a x bμ
    = nμ

    ∴[itex](a χ u)/n[/itex] = μ
    [itex](a χ u)/(a χ b)[/itex] = μ

    The same can be applied for obtaining an equation with λ. The problem arises out of the fact that I wouldn't know how to incorporate the '.n' portion to the numerator and denominator (though it seems quite obvious that their cancellation would yield the above equation, it must serve some purpose).

    2. I can't find any sort of methodology that would yield a vector, b', perpendicular to a in P1 when only given the information above. I would assume it has something to do with question 1 (I could be wrong), though I can't see how they relate to one another.


    Thanks in advance for any help.
     
  2. jcsd
  3. Nov 1, 2012 #2

    Mark44

    Staff: Mentor

    Mod note: I tweaked what you wrote to make it more readable. In particular, I replaced the character you used for the cross product (Greek letter chi?) with X.
    What is n?
     
  4. Nov 1, 2012 #3

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    But of course, you can't divide by a vector. What you do have correct is$$
    a\times u =\mu(a\times b)$$with vectors on both sides. You can dot both sides of that equation with ##n##, giving$$
    (a\times u)\cdot n = \mu(a\times b) \cdot n$$Now this is a scalar equation which you can solve for ##\mu##. Do you see why ##(a\times b) \cdot n\ne 0\ ##, allowing you to divide by it?

    And you do the similar thing to solve for ##\lambda##.
    For 2 think about where the vector ##(a \times b)\times a## would be.
     
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