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    Got a little question, ive done it but my way is a little long winded

    L and M are 2i-3j+3k and 5i+j+ck respectively. The point N is such that OLMN is a rectangle!!!

    The way i have done it is to calculate the vector equations of 0L and LM and use a.b=0. i then some how come to (c-5)=-22/3t

    the answer is c=5.

    THIS IS JUST WAY TO LONG TO BE THE RIGHT WAY

    any help will be appreciated
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    What is the actual question? What am i trying to find out?
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    trying to calculate the value of c
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    (Original post by kashif_k)
    Got a little question, ive done it but my way is a little long winded

    L and M are 2i-3j+3k and 5i+j+ck respectively. The point N is such that OLMN is a rectangle!!!

    The way i have done it is to calculate the vector equations of 0L and LM and use a.b=0. i then some how come to (c-5)=-22/3t

    the answer is c=5.

    THIS IS JUST WAY TO LONG TO BE THE RIGHT WAY

    any help will be appreciated
    You can say that the direction vectors \vec{OL} and \vec{MN} are equal in magnitude and direction.
    \vec{OL}=\vec{MN}\\

\vec{OL}=\vec{ON}-\vec{OM}\\

\vec{ON}=\vec{OL}+\vec{OM}
    Are you sure you don't have some other information such as the position vector of N?
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    i know its gonna be something really easy that im over complicating
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    (Original post by Gaz031)
    You can say that the direction vectors \vec{OL} and \vec{MN} are equal in magnitude and direction.
    \vec{OL}=\vec{MN}\\

\vec{OL}=\vec{ON}-\vec{OM}\\

\vec{ON}=\vec{OL}+\vec{OM}
    Are you sure you don't have some other information such as the position vector of N?
    Nope thats all they give! the question can be found on page 78 Q9 in the c4 heinemenn text book!
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    (Original post by kashif_k)
    Nope thats all they give! the question can be found on page 78 Q9 in the c4 heinemenn text book!
    Your method is right.

    The answer in the back is also right.

    I am typing up my working to word and will post it here within 5 mins.
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    Solution is enclosed.
    Attached Files
  1. File Type: doc vectors.doc (23.0 KB, 100 views)
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    i like the lil solution by samydavyson at the end lol...btw on ure revision sheet for C4 there diesnt seem ot b nething on the trapezium rule...that sheet is HEAVY
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    Thanks mate. I realised my mistake afte posting my last message. Basically when doing the a.b=o thing instead of using the direction vector bits of the equations of the ;lines i was using the whole thing!! DOH
    I checked my workings with yours and its the same

    thanks again mate (by the way im now stuck on differntial equations- check the other thread
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    Just been looking at samy's solution, and the bit at the end:

    l.p = 0 = (2x3) + (-1x4) + (3 * (c-3) = 6 - 4 + 3(C-3)
    2+3C-9
    3C - 7 = 0
    3C = 7
    C = 7/3

    How do you get -1 * 4 = -12, or am I missing something?

    Edit: plus, on p = m-l, there's (1) - (-1) = 4?!


    Edit: just worked it out on paper, and it came out fine (C=5), so I don't know what happened in that word file? :confused:
 
 
 
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