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    Can someone help me with question 7c. It's from solomon sheets on vectors.Name:  image.jpg
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    (Original post by ACarter8)
    Can someone help me with question 7c. It's from solomon sheets on vectors.Name:  image.jpg
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    I haven't actually done it, but this is what I would do, from the equation of line l2 we can form a position vector for R in terms of t. As you have worked out the position vectors of P and Q, you can work out the distance. As we know that PQ=QR, work out the vector QtoR, this will of course be in terms of t. Then work out the distance, however this distance will be equal to PQ, then solve for 2, if will be a quadratic so you should get two values (hopefully this is right!)

    EDIT: Just realized that a quicker method is just to use the direction vector of l2 and see what multiple it is of the distance PQ, and then do +/- of that.
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    (Original post by randlemcmurphy)
    I haven't actually done it, but this is what I would do, from the equation of line l2 we can form a position vector for R in terms of t. As you have worked out the position vectors of P and Q, you can work out the distance. As we know that PQ=QR, work out the vector QtoR, this will of course be in terms of t. Then work out the distance, however this distance will be equal to PQ, then solve for 2, if will be a quadratic so you should get two values (hopefully this is right!)

    EDIT: Just realized that a quicker method is just to use the direction vector of l2 and see what multiple it is of the distance PQ, and then do +/- of that.
    The ms was your edited answer, still don't get it really.

    EDIT: lol it's just clicked. Thanks for the help
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    Can someone help me with this question please????
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    (Original post by ridirahman)
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    Can someone help me with this question please????
    Two approaches:
    1. Use the double angle formula multiple times and expand out. Long and arduous?

    Use these sneaky factor formula that a lot of schools omit from teaching.



    Try the sin A - sin B one.
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    (Original post by Alex:)
    Two approaches:
    1. Use the double angle formula multiple times and expand out. Long and arduous?

    Use these sneaky factor formula that a lot of schools omit from teaching.



    Try the sin A - sin B one.
    There is a far cleverer way of doing this ...
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    (Original post by TeeEm)
    There is a far cleverer way of doing this ...
    I know it... Hehe.


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    (Original post by physicsmaths)
    I know it... Hehe.


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    I should hope so, otherwise no Cambridge ...
    You will end up going to UCL like me since Imperial also rejected you ...
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    (Original post by TeeEm)
    There is a far cleverer way of doing this ...
    I don't see a quicker method of approaching this without using the factor formulae. Care to share?

    EDIT: Is it something obvious like taking one sin to the other side and inverse sin both sides?
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    (Original post by Alex:)
    I don't see a quicker method of approaching this without using the factor formulae. Care to share?

    EDIT: Is it something obvious like taking one sin to the other side and inverse sin both sides?
    kind of if it is done properly ...
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    (Original post by TeeEm)
    kind of if it is done properly ...
    Rearrange
    sin5x = -sinx

    Since sin is an odd function:
    sin5x = sin-x

    +2kpi and (2k+1)pi need to be considered

    To be honest, the factor formulae are probably easier, as you have to consider the periodic nature of the sin graph as well as the rotational symmetry of the sin graph.
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    (Original post by Alex:)
    Rearrange
    sin5x = -sinx

    Since sin is an odd function:
    sin5x = sin-x

    +2kpi and (2k+1)pi need to be considered

    To be honest, the factor formulae are probably easier, as you have to consider the periodic nature of the sin graph as well as the rotational symmetry of the sin graph.
    It depends on your level ...
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    (Original post by TeeEm)
    It depends on your level ...
    That's true. I've always found it very dirty to be using inverse trig functions when both sides have trig functions. I am probably one of the very few people who use the factor formulae.
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    (Original post by TeeEm)
    I should hope so, otherwise no Cambridge ...
    You will end up going to UCL like me since Imperial also rejected you ...
    Brutal AF
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    (Original post by Brubeckian)
    Brutal AF
    ..
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    (Original post by Alex:)
    That's true. I've always found it very dirty to be using inverse trig functions when both sides have trig functions. I am probably one of the very few people who use the factor formulae.
    I would agree that you must use a technique appropriate to the standard of the question and never show off...
    There are only marks for correct answers (no marks for bravery unfortunately)
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    (Original post by TeeEm)
    I should hope so, otherwise no Cambridge ...
    You will end up going to UCL like me since Imperial also rejected you ...
    F imperial lolz. Na ill end up at warwick hehe


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    (Original post by physicsmaths)
    F imperial lolz. Na ill end up at warwick hehe


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    Warwick....
    I would not have gone there even if you paid me ...
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    (Original post by TeeEm)
    Warwick....
    I would not have gone there even if you paid me ...
    Why not lol?



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    (Original post by physicsmaths)
    Why not lol?



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    personal dislike of the place
 
 
 
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