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    (for reference rev ex2, p3 green heinman book....qu.33)
    Referred to a fixed origin O, the point A and B have posn vectors (5i-j-k) and (i-5j+7k) respectively.
    Find the posn vector of the point D, where D is not A, on AB such that OD=OA

    Okay so if i know D is on AB, it must satisfy 5i-j-k+t(-4i-4j+8k).
    But , if OD=OA, then 5i-j-k+t(-4i-4j+8k)=(5i-j-k), but that just gives me D=A[cuz t=o after coefficients are equated], and qu says this is not the case, so where am i going wrong? :confused:
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    (Original post by BossLady)
    (for reference rev ex2, p3 green heinman book....qu.33)
    Referred to a fixed origin O, the point A and B have posn vectors (5i-j-k) and (i-5j+7k) respectively.
    Find the posn vector of the point D, where D is not A, on AB such that OD=OA

    Okay so if i know D is on AB, it must satisfy 5i-j-k+t(-4i-4j+8k).
    But , if OD=OA, then 5i-j-k+t(-4i-4j+8k)=(5i-j-k), but that just gives me D=A[cuz t=o after coefficients are equated], and qu says this is not the case, so where am i going wrong? :confused:
    When it says OA = OD it means (I think) that they have the same modulus!
    So the triangle OAD is an isoceles triangle, with (lemgth of) OA = (length of) OD , if that helps
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    Yeh, just checked it - compare the modulus of OA, as given, and as from the formula you worked out.
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    (Original post by Fermat)
    When it says OA = OD it means (I think) that they have the same modulus!
    So the triangle OAD is an isoceles triangle, with (lemgth of) OA = (length of) OD , if that helps
    ohhh lol that's why the book wrote it as |OA|=|OD|!!! D'oh!
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    ok soo if i write sqrt(27)=5i-j-k+t(-4i-4j+8k), ah wait xi+yj+zk=5i-j-k+t(-4i-4j+8k).
    i: x=5-4t
    j:y=-1-4t
    k:z=-1+8t,

    so (5-4t)i+(-1-4t)j+(-1+8t)k=sqrt(27)

    Am i on the right track or am i just going round in circles here?i suspect the latter
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    (Original post by BossLady)
    ok soo if i write sqrt(27)=5i-j-k+t(-4i-4j+8k), ah wait xi+yj+zk=5i-j-k+t(-4i-4j+8k).
    i: x=5-4t
    j:y=-1-4t
    k:z=-1+8t,

    so (5-4t)i+(-1-4t)j+(-1+8t)k=sqrt(27)

    Am i on the right track or am i just going round in circles here?i suspect the latter
    Almost!
    It should be modulus(lhs) = sqrt(27)
    then you get a quadratic in t.
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    (Original post by Fermat)
    Almost!
    It should be modulus(lhs) = sqrt(27)
    then you get a quadratic in t.
    oooo 3i-3j+3k!!
    the book answer! yay! thanks fermat!
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    (Original post by BossLady)
    oooo 3i-3j+3k!!
    the book answer! yay! thanks fermat!
    fermat strikes again.
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    (Original post by Fermat)
    Almost!
    It should be modulus(lhs) = sqrt(27)
    then you get a quadratic in t.
    what's lhs?
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    (Original post by wibble...)
    what's lhs?
    Left Hand Side?

    ~Darkness~
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    (Original post by Darkness)
    Left Hand Side?

    ~Darkness~
    yep
 
 
 
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