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    The position vectors of the points A and B relative to the origin, O, are

    5i+ 4j+ k

    -i+j-2k respectively

    Find the position vector of the point p which lives on AB such that AP= 2BP
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    (Original post by APersonYo)
    The position vectors of the points A and B relative to the origin, O, are

    5i+ 4j+ k

    -i+j-2k respectively

    Find the position vector of the point p which lives on AB such that AP= 2BP
    Find the vector A->B.
    Divide it by three, and add it to A.
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    (Original post by carpetguy)
    Find the vector A->B.
    Divide it by three, and add it to A.
    Thank you xo
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    (Original post by APersonYo)
    Thank you xo
    Do you understand why? I hope you do. Let me know if you don't.

    Tbf it requires basic vector understanding and I would assume your teacher is not providing this.
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    (Original post by carpetguy)
    Do you understand why? I hope you do. Let me know if you don't.

    Tbf it requires basic vector understanding and I would assume your teacher is not providing this.
    Erm, Well, I did what you said, and its not right:

    Vector AB= (-1,1,-2) - (5,4,1) = (-6,-3,-3)

    (-6,-3,-3) divided by 3= (-2,-1,-1)

    O to P= ( 5,4,1)+ (-2,-1,-1)= (3,3,0)
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    disclaimer: not completely sure
    but if AP=2BP and P is a vector xi+yj+zk then x-A(i)=2(x-B(i)), y-A(j)=2(y-B(j))
    and z-A(k)=2(y-B(k)). you can now solve for x y and z where A(i) is the component of the vector A in the i direction, A(k) is the component of the vector A in the k direction, etc (same idea for B).
    Could you tell me if this gets you the right answer or not? I'm also learning
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    What I think:

    AP = 2BP

    Therefore: p - a = 2(p - b)
    p - a = 2p - 2b
    p = 2b - a

    Therefore: p = 2(-1, 1, -2) - (5, 4, 1)
    p = (-2,2,-4) - (5,4,1)
    p = (-7, -2, -5)
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    (Original post by APersonYo)
    Erm, Well, I did what you said, and its not right:

    Vector AB= (-1,1,-2) - (5,4,1) = (-6,-3,-3)

    (-6,-3,-3) divided by 3= (-2,-1,-1)

    O to P= ( 5,4,1)+ (-2,-1,-1)= (3,3,0)

    Sorry. I skimmed the question. You should find 2 thirds of the vector A->B and add that to A.

    So multiply (-2,-1,-1) by 2 and then add the result to A.

    A suggestion. Get rid of the third dimension and draw this on a graph so you can appreciate the solution. Do you know where the factor of 3 is coming from?
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    (Original post by Daiblain)
    disclaimer: not completely sure
    but if AP=2BP and P is a vector xi+yj+zk then x-A(i)=2(x-B(i)), y-A(j)=2(y-B(j))
    and z-A(k)=2(y-B(k)). you can now solve for x y and z where A(i) is the component of the vector A in the i direction, A(k) is the component of the vector A in the k direction, etc (same idea for B).
    Could you tell me if this gets you the right answer or not? I'm also learning
    This gets confusing as you are finding vectors which are head on, so some of your negatives will be confused with positives.


    (Original post by Nikita_2313)
    What I think:

    AP = 2BP

    Therefore: p - a = 2(p - b)
    p - a = 2p - 2b
    p = 2b - a

    Therefore: p = 2(-1, 1, -2) - (5, 4, 1)
    p = (-2,2,-4) - (5,4,1)
    p = (-7, -2, -5)
    Note that the question's statement of AB and BP does not refer to vectors, simply their magnitude. So, to keep this all working, you'd want to do b-p rather than p-b.


    Both of you guys, all this algebra is fine, but really is unnecessary, if I am correct in my method.
 
 
 
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