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Vector linear relationship

  1. Apr 29, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the general solution to the equation: ##{u}\times{(i+4j)}=3k##
    2. Relevant equations
    ##u=(ai+bj+ck)##

    3. The attempt at a solution
    I am having trouble visualising what to do here.

    So my throught so far is that, if i do ##(ai+bj+ck)\times {(i+4j)}=(0i-0j+(4a-b)k) ##
    now if I equate ##(0i-0j+(4a-b)k)=(0i+0j+3k)## I get ##4a-b=3## and this is where i draw a blank,it's the general equation part that I dont get.

    I mean if I rearange I could do for a or b, I preffer b as it have no fractions, ##4a-3=b##, so I assume that I would sub this back into u which would give me ##u=(0i+(4a-3)j+0k)##. Now doe the general equation, mean vector equation of the line. Normally I would find the direction vector and the starting vector but it being a cross product has thrown me a bit. Am I on the right lines or have I gone very wrong?
     
  2. jcsd
  3. Apr 29, 2017 #2

    jedishrfu

    Staff: Mentor

    From the first you know that u is perpendicular to k hence u can be written as ai + bj and from there sub into the cross product and solve it.
     
  4. Apr 29, 2017 #3
    Hi, is that not what I have done, the ans in the back of the exercsie book gives the ans as: ##u=-3j+t(i+4j)## where has the parameter t come into, p.s I did not look at the ans before I posted, but still unawre how u can equal this, as to me it look like the vector equation of a line.
     
  5. Apr 29, 2017 #4
    You just need to equate magnitudes. since any two dimensional vector have $k$ as the unit vector of vector product.
     
  6. Apr 29, 2017 #5
    Ok, I am really not understanding here, how dose equating the magnitudes help with this problem as there is a parameter t. Could you please expand, if you calculate the magnitude surely you just working out the area of a parallelogram?
     
  7. Apr 29, 2017 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    ##\vec u=a \vec i + b \vec j +c \vec k##. You obtained b=4a-3 which is correct, and a is arbitrary. For the general solution, you can not take the coefficient of ##\vec i ## zero in vector ##\vec u##.
     
  8. Apr 29, 2017 #7
    so from what u are saying is that ##a=t, u=ai+bj=ti+(4t-3)j## now I assume that this is t can be any number along the u vector?
     
  9. May 1, 2017 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

    t is an arbitrary number, and u depends on it. "number along the u vector" has no sense.
     
  10. May 3, 2017 #9
    Thank you for you help after taking a step back I relieased what was going on. Once again thank you for the help
     
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