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    With respect to an origin O, the position vectors of the points L and M are 2i - 3j + 3k and 5i + j + ck respectively, where c is a constant. The point N is such that OLMN is a rectangle.

    a) Find the value of c

    b) Write down the position vector of N

    c) Find, in the form r = p + tq, an equation of the line MN

    Cant do any of it, can anyone give me hand?
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    help, please, bump, bla bla bla
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    OL and ON are perpendicular, so dot product is 0. Solve to find c

    The line MN is
    r = M + t(-OL)
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    (Original post by imasillynarb)
    With respect to an origin O, the position vectors of the points L and M are 2i - 3j + 3k and 5i + j + ck respectively, where c is a constant. The point N is such that OLMN is a rectangle.

    a) Find the value of c

    b) Write down the position vector of N

    c) Find, in the form r = p + tq, an equation of the line MN

    Cant do any of it, can anyone give me hand?

    What are the answers?
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    (Original post by Silly Sally)
    What are the answers?
    a) Find c:

    LM = 3i + 4j + (c-3)k
    OL = 2i - 3j + 3k

    these are perpendicular.. therefore (2i - 3j + 3k)(3i + 4j + (c-3)k) = 0

    6 - 12 + 3(c-3) = 0

    c = 15/3 = 5

    c = 5, correct?
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    (Original post by Sahir)
    a) Find c:

    LM = 3i + 4j + (c-3)k
    OL = 2i - 3j + 3k

    these are perpendicular.. therefore (2i - 3j + 3k)(3i + 4j + (c-3)k) = 0

    6 - 12 + 3(c-3) = 0

    c = 15/3 = 5

    c = 5, correct?
    Yes
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    (Original post by Sahir)
    a) Find c:

    LM = 3i + 4j + (c-3)k
    OL = 2i - 3j + 3k

    these are perpendicular.. therefore (2i - 3j + 3k)(3i + 4j + (c-3)k) = 0

    6 - 12 + 3(c-3) = 0

    c = 15/3 = 5

    c = 5, correct?
    Thanks - how do you do part b?
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    (Original post by imasillynarb)
    Yes
    Heh i'm glad, cuz i gave up with the rest! to be honest with you i've not seen any questions on P3 which are GCSE-style vectors questions like in ex.5A of that Heinemann text book, do you know what i mean?
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    Cant do b) help!!

    N = xi + yk + zk

    I did, ON is perpendicular to OL

    Meaning 2x - 3y + 3z = 0

    Now Im stuck..
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    (Original post by imasillynarb)
    Cant do b) help!!

    N = xi + yk + zk

    I did, ON is perpendicular to OL

    Meaning 2x - 3y + 3z = 0

    Now Im stuck..
    PLEASE give me the answers - sometimes i work back from ans
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    (Original post by Silly Sally)
    PLEASE give me the answers - sometimes i work back from ans
    5, 3i + 4j + 2k
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    ON = OM - OL = (3, 4, 2)
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    b) Write down the position vector of N

    The position vector of N is ON, and this is equal to LM. This is just OM - OL.

    so this gives (3, 4, 2)



    c) Find, in the form r = p + tq, an equation of the line MN


    this is just:

    OM + t(OL)

    as OM is a vector onto the line,

    and OL is a vector parallel.

    Hope this helps.
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    Yeh ive done it now, cheers everyone
 
 
 

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