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Solution to vector equation involving vector product

  1. Mar 22, 2015 #1
    I want to find the solution of vector X. I am using text from Alan F. Beardon Algebra and Geometry as attached. I dont know how the solution is derived for the following equation.

    ## x + (x × a) = b ##

    The second solution when ## a \times b \neq 0 ## then X cannot be b. Is it possible to factorize x out from the Left-hand-side of the above equation?
     

    Attached Files:

  2. jcsd
  3. Mar 22, 2015 #2

    Mark44

    Staff: Mentor

    It seems to me that you're missing the point, which is not to solve the equation, but only to determine whether it has one or more solutions.

    The assumption is that you have two solutions -- let's call them x1 and x2. How they were obtained is not important. In the page whose image you included, they assume that y is the difference of any two solutions, so y = x1 - x2. Continue from here.
     
  4. Mar 22, 2015 #3

    mathman

    User Avatar
    Science Advisor
    Gold Member

    When you expand the vector equation as 3 scalars in terms of the components of x, you will have 3 linear equations in 3 unknowns. There is no way to factor out x.
     
  5. Mar 22, 2015 #4
    For solving vector equations like the one you got, there is a method. Since a,b and axb are 3 non-coplanar vectors, you can take
    ##x=ma+nb+p(a*b)##
    Just like you take x=ai+bj+ck.
    Where m,n,p are scalars.

    Try finding the values of m,n,p by substituting the value of x in the given vector equation.

    I am sorry @Mark44 for providing the solution.
     
  6. Mar 22, 2015 #5

    WWGD

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    There is no natural multiplication of vectors in ## \mathbb R^3 ## that allows you to factor.
    EDIT I meant other than the obvious dot and cross products; then some expressions containing
    both may be at times factorable, but not otherwise.
     
  7. Mar 23, 2015 #6
    Hey @AdityaDev, A & B are not necessarily orthogonal so it is not exactly like ai + bj + ck where i, j & k are mutually orthogonal! so??
     
  8. Mar 23, 2015 #7
    It doesn't matter. I have solved many questions using this method.
    In certain questions, you have to use logic.
    Example: Solve $$ \vec{r}\times\vec{b}=\vec{a}\times\vec{b}$$
    Given r.c=0
     
  9. Mar 24, 2015 #8
    Aditya's missing explanations in the above post can be found in this attachment!!

    Why do we want to find the vector X with basis ## a, b, a \times b ## ?

    Why not in simple ##i, j, k##? Is it because it is simplest and cancels out easily in the equation ##x + (x \times a) = b##

    Then why is each component ## x_1 , x_2 , x_3 ## not divided by the scalar triple product ## [a, b, a \times b] ## e.g.

    ## X = [X, b, a \times b]/[a, b, a \times b]*x_1 + [a, X, a \times b]/[a, b, a \times b]*x_2 + [a, b,X]/[a, b, a \times b]*x_3 ##
     

    Attached Files:

    Last edited: Mar 24, 2015
  10. Mar 24, 2015 #9
    It was a question for you to practice. I already knew the solution.
     
  11. Mar 24, 2015 #10
    I have attached my attempt at finding ##x_1 , x_2, x_3##. How do I simplify further??
     

    Attached Files:

  12. Mar 24, 2015 #11
    What you have done is correct.
    Capture.PNG

    Now, expand (axb)xa.
    (axb)xa=(a.a)b-(a.b)a ---> Vector triple product.
    Now you will get :

    Capture2.PNG
    Then expand the last term to get ##X_3|a|^2\vec{b}-X_3(a.b)\vec{a}##
    Now you will get this expression :
    Capture3.PNG
    then group axb and bxa terms as you can express axb as -bxa.
    then group the terms with vector a.
    also group the terms of vector b.
    So you have three terms.
    Then equate the coefficients of all vectors other than b like a and axb to find x1,x2,x3.
     
  13. Mar 25, 2015 #12
    Hey @AdityaDev, thanks for your help. I was not that familiar with vector triple product so I could not spot the expansion of ##a \times b \times a ## into (a.a)b - (a.b)a.
     
  14. Mar 25, 2015 #13
    EDIT: equate the coefficients of all vectors other than b "to zero". Now you can find x1, x2, x3.
    Did you get the correct answer?
     
  15. Mar 25, 2015 #14
    First take the cross product with a on both sides,
    This should give you , (b-x) + (a.x)a - (|a|^2)x = b×a.......(equation 1)here I have substituted x × a as b-x,
    Substitute value of a.x by using a second equation , in which you take dot product on both sides of the original equation , x.a + (x×a).a= b.a
    Which gives x.a + 0 = b.a
    Now substitute a.x as b.a

    This should allow you to collect terms of x.
     
    Last edited: Mar 25, 2015
  16. Mar 25, 2015 #15
    You will get (a.x)a=(b-x)×a + kx
    also, you cant find a.x from this equation because it is grouped with a vector and if you thought this: $$a.x=\frac{(b-x)\times\vec{a}+kx}{\vec{a}}$$
    Then you are wrong.
    You can however find the value of vector a.
     
  17. Mar 25, 2015 #16
    Please see the edited post above .
     
  18. Mar 26, 2015 #17
    Hi, can you all answer my questions??

    Why do we want to find the vector X with basis ##a, b, a \times b## ?

    Why not in simple ##i,j,k##? Is it because it is simplest and cancels out easily in the equation## x+(x×a)=bx + (x \times a) = b##

    Then why is each component ## x_1 , x_2 , x_3## not divided by the scalar triple product ##[a, b, a \times b]## e.g.

    ## X=[X,b,a \times b]/[a,b,a \times b]∗x_1+[a,X,a \times b]/[a,b,a \times b]∗x_2+[a,b,X]/[a,b,a \times b]∗x_3 ##

    Thanks!!
     
    Last edited: Mar 26, 2015
  19. Mar 27, 2015 #18
    a, b, and a×b are a set of non coplanar vectors and any vector in three dimensional space can be expressed as a linear combination of three non-coplanar vectors,( just like i, j, k)
     
  20. Mar 28, 2015 #19
    Hey @Ananya0107, Can you explain why was a, b and a x b selected why not any other vector as you say " any vector in three dimensional space can be expressed as a linear combination of three non-coplanar vectors,( just like i, j, k)"?

    Thanks!
     
  21. Mar 30, 2015 #20
    The question contained the vectors a and b so it is easier to express the required vector x in terms of a , b and a×b , it is just easier for computation .
     
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