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    When a vector is parallel to the (i-j) vector I tried to put the i part minus the j part and then equate everything to 0. This results in a wrong answer and I don't understand why..
    Like looking at the ms/solution it equates the i part to the NEGATIVE j part. Why is that?
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    (Original post by RK1998)
    When a vector is parallel to the (i-j) vector I tried to put the i part minus the j part and then equate everything to 0. This results in a wrong answer and I don't understand why..
    Like looking at the ms/solution it equates the i part to the NEGATIVE j part. Why is that?
    post a question and someone would help you
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    (Original post by TeeEm)
    post a question and someone would help you
    a= (5t-3)i + (8-t)j ms**

    when t=0, v=(2i-5j) ms*

    a) velocity after t seconds is (2.5t^2 -3t +2)i + (8t-0.5t^2 -5)j

    b) Find the value of t for which is moving parallel to i-j

    c) Find the speed when it is moving parallel to i-j


    # I need help with part b please #
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    (Original post by RK1998)
    a= (5t-3)i + (8-t)j ms**

    when t=0, v=(2i-5j) ms*

    a) velocity after t seconds is (2.5t^2 -3t +2)i + (8t-0.5t^2 -5)j

    b) Find the value of t for which is moving parallel to i-j

    c) Find the speed when it is moving parallel to i-j


    # I need help with part b please #
    When it's moving parallel to i - j, the general velocity vector will be equal to some constant multiplied by the vector I - j.
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    (Original post by RK1998)
    Yeah but does this mean that I equate to 0 the velocity of i+j or i-j ?
    When P moving parallel to \[\begin{pmatrix} \ 1, -1\end{pmatrix}\] it means that v = \lambda \[\begin{pmatrix} \ 1, -1\end{pmatrix}\].
 
 
 
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