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    this question is not tough,i think.
    I just don't know how to solve it
    WOuld u guys help me out

    O is the origine.
    OA=i + 2j
    OB= -i+4j

    1/2^0.5(i+j) is a unit vector perpenducular to AB

    find the position vector of the foot of the perpendicular from the origin to the line segment AB.


    2.

    position vectors of A ,B,C,D and E are a , b,c,d,e respectively.
    A is the midpt of OB and E divides AC in the ratio 1:2.
    If e=1/3 d , show that OCDB is a parallelogram..




    Thank you very much ...
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    (Original post by Fong_Leonard)
    this question is not tough,i think.
    I just don't know how to solve it
    WOuld u guys help me out

    O is the origine.
    OA=i + 2j
    OB= -i+4j

    1/2^0.5(i+j) is a unit vector perpenducular to AB

    find the position vector of the foot of the perpendicular from the origin to the line segment AB.

    Thank you very much ...
    find AB = OB - OA

    The line going through the origin perp to AB is r = t/2^0.5(i+j) for a scalar t. Find the value of t for which the two lines cross
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    i don;t quite get it . why do u introduce the t .
    and which equation do u compare to find out the value of t??
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    (Original post by Fong_Leonard)
    i don;t quite get it . why do u introduce the t .
    and which equation do u compare to find out the value of t??
    The eqn of the line through AB is,
    r1 = OA + λ(OB - OA)

    The line passing thro' the origin parallel to the unit vetor v (let v=(1/√2)(i+j) ) is r2 = tv, say.
    When r1 and r2 intersect, they are equal, so put r1=r2, and solve for λ and t.

    I got (3/2)(i+j)
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    (Original post by Fong_Leonard)
    .....

    2.

    position vectors of A ,B,C,D and E are a , b,c,d,e respectively.
    A is the midpt of OB and E divides AC in the ratio 1:2.
    If e=1/3 d , show that OCDB is a parallelogram..




    Thank you very much ...
    AC = c - a
    OE = OA + (1/3)AC ( E divide AC in ratio 1:2)
    OE = a + (1/3)(c - a)
    OE = c/3 + 2a/3
    ==========

    OD = 3OE, (e = d/3)
    OD = c + 2a
    CD = OD - OC
    CD = c + 2a - c
    CD = 2a
    but,
    OB = 2a (A is mid-pt of OB)
    => CD parallel to OB
    =============

    BD = OD - OB
    BD = c + 2a - 2a
    BD = c
    but,
    OC = c
    => BD parallel to OC
    =============

    => OCBD is a parallelogram
    ==================
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