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Vector Equations

  1. Nov 5, 2005 #1
    Hi all,

    I would like some help with a tricky vector equation, I need to solve for x.

    2x-(xdoti)i = i + 5j - (xi)

    The dot is a dot product and not multiply. i am not sure how to deal with the dot product, I am little confused. Any help would be excellent. Thanks. :smile:
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  3. Nov 5, 2005 #2


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    [tex]\vec{x}=x_i \hat{i} + x_j \hat{j}[/tex]
    [tex]\vec{y}=y_i \hat{i} + y_j \hat{j}[/tex]
    [tex]x \dotprod y = x_i y_i + x_j y_j[/tex]
  4. Nov 5, 2005 #3


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    I'm a little confused myself! (xdoti) implies that x is a vector but then "xi" is not clear. If x= ai+ bj then "x dot i" is just a so "2a- (x dot i)i would be
    2ai+ 2bj- ai= ai+ 2bj. But then I have no idea what (xi) means.
  5. Nov 5, 2005 #4
    The (xi) at the end is actually x cross i, I didnt read it carefully emough. Now, I have had another crack at it working out the dot and cross product sections in the equation, this is my working:

    2x-(ai + bj).(i) = i +5j -((ai + bj) cross i)

    substituting the x for ai(unit vector) and bj(unit vector)

    2x-ai = i + 5j - bk
    x = [(a+1)i + 5j -bk]/2

    is this working correct? The bit I am most unsure about is the working out of the cross product using matrices. Thanks for the help, sort of clicked my brain into gear.
  6. Nov 5, 2005 #5


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    Write x as ai+ bj+ ck. Then 2x= 2ai+ 2bj+ 2ck and (x dot i)i= ai so 2x- (x dot i)i= (2a-a)i+ 2bj+ 2ck= ai+ 2bj+ 2ck. x cross i= cj- bk.

    2x-(xdoti)i = i + 5j - (x cross i) is
    ai+2bj+ 2ck= i+ (5-c)j- bk

    We must have a=1, 1b=5- c and 2c= -b.

    Can you solve those?
  7. Nov 6, 2005 #6
    Thanks guys, got it all sorted I think. Realised my mistake with the matrix. Got x to be: (i + 2j + k) which when substitutued back into the original formula all works out. Thanks again. :biggrin:
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