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Vector components hard!

  1. Sep 3, 2005 #1
    hi all got some problems with vectors so waswondering if anyone can check what i've done
    is correct thanks

    find the vector component of 2i+j-k in the direction of i-3j+2k
    basically i use the rule
    w^= i-3j+2k / sqrt(14)

    thus v=2i+j-k
    therefore a=(2i+j-k)*( i-3j+2k / sqrt(14) ) * (i-3j+2k / sqrt(14))
    a= [2(1)-3(1)-1(2) /sqrt(14)] * [i-3j+2k / sqrt(14)]
    a= -3/sqrt(14)* i-3j+2k / sqrt(14)

    thus the vector componentis a=-3/14( i-3j+2k) ???

    2. find the parametric equationsof the straightline of intersection of the planes

    i used gaussian elimination and got the tableu
    1 -2 3 | 5
    3 1 -2 | 1 R2-3R1

    1 -2 3 | 5
    0 7 -11| -14


    sub z=t into 7y-11z=-14
    y=11t/7 - 2

    sub y=11t/7 - 2 into x-2y+3z=5
    x= 5 + 1/7t

    therefore are the parametric equations
    x= 5 + 1/7t
    y=11t/7 - 2

    thanks all
  2. jcsd
  3. Sep 3, 2005 #2


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    Homework Helper

    1. You found the projection vector along w, but it sounds like they just wanted the length of this vector. That is, just [itex]\vec v \cdot \hat w[/itex]. Think of it like this: if you rearranged your axes so that w pointed along the new x axis, what would be the new x component of v?

    2. The easiest way to check this is to plug x, y, and z back into the two plane equations and see if they solve them for all t (ie, t drops out and you get something like 1=1).
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