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    So I've been scratching my head over this - I have had half a lesson on vectors and I'm still in that stage of just starting to get it, and then this threw me. I went to my college maths lunchtime session for help and the only free teacher couldn't do it either (I spent 25 minutes sitting there while she drew out what I'd drawn, then stared at it helplessly), but it's my homework so I feel obliged to understand it and get it done! Any help would be greatly appreciated!

    L and M are the points (1/3, -1, 2) and (1, 5, 6) respectively. The position vector of point P to relative to the origin O is i + 3j + 6k. Point R lies on OP such that OR:RP = K:1. Find k if R, L and M are collinear.

    ??????:confused:
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    (Original post by DayFlower)
    So I've been scratching my head over this - I have had half a lesson on vectors and I'm still in that stage of just starting to get it, and then this threw me. I went to my college maths lunchtime session for help and the only free teacher couldn't do it either (I spent 25 minutes sitting there while she drew out what I'd drawn, then stared at it helplessly), but it's my homework so I feel obliged to understand it and get it done! Any help would be greatly appreciated!

    L and M are the points (1/3, -1, 2) and (1, 5, 6) respectively. The position vector of point P to relative to the origin O is i + 3j + 6k. Point R lies on OP such that OR:RP = K:1. Find k if R, L and M are collinear.

    ??????:confused:
    have you done equations of lines?
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    (Original post by DayFlower)
    So I've been scratching my head over this - I have had half a lesson on vectors and I'm still in that stage of just starting to get it, and then this threw me. I went to my college maths lunchtime session for help and the only free teacher couldn't do it either (I spent 25 minutes sitting there while she drew out what I'd drawn, then stared at it helplessly), but it's my homework so I feel obliged to understand it and get it done! Any help would be greatly appreciated!

    L and M are the points (1/3, -1, 2) and (1, 5, 6) respectively. The position vector of point P to relative to the origin O is i + 3j + 6k. Point R lies on OP such that OR:RP = K:1. Find k if R, L and M are collinear.

    ??????:confused:
    I get k=2
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    (Original post by TenOfThem)
    I get k=2
    I get k= 2 also (but black marks for whoever decided to use k [or K] as a constant of proportionality as well as k as a basis vector!).

    Not a very pretty question
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    (Original post by davros)
    I get k= 2 also (but black marks for whoever decided to use k [or K] as a constant of proportionality as well as k as a basis vector!).

    Not a very pretty question
    Yeah the use of k annoyed me too

    Was better once I drew a diagram
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    (Original post by TenOfThem)
    Yeah the use of k annoyed me too

    Was better once I drew a diagram
    I can't even draw diagrams when it gets to 3 dimensions!

    I just wrote down all the relationships I could think of including one for a general point on the line LM extended and then slogged my way through the algebra. Probably not the most efficient method :confused:
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    (Original post by DayFlower)
    L and M are the points (1/3, -1, 2) and (1, 5, 6) respectively. The position vector of point P to relative to the origin O is i + 3j + 6k. Point R lies on OP such that OR:RP = K:1. Find k if R, L and M are collinear.
    The key is to note that R,L and M are collinear i.e. that they sit on the same line, but also that O, R and P are collinear i.e they sit on another line. Since R is a point in both lines, these two lines intersect at R.

    You can then find an equation for line RLM (you have two points on this line, namely L and M) and an equation for line ORP (you have two points on this line, namely O(0,0,0) and P), then you find their point of intersection (which is at R) via the usual techniques.

    It's then easy to find the required ratio.
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    wow one question and 4 keen helpers!...
    the funny thing is the person with the actual question is too intimidated from the heavy presence ... of the good, the bad, the ugly and the pain in the rear ( I am not implying in respective order )




    I agree that the atsruser's method is the simplest and that is why I was asking whether the OP is aware of lines The question can be done without it although it is conceptually harder.

    I have a very similar question (Q118) in the link

    http://www.madasmaths.com/archive/ma...ons_part_a.pdf
    (longish download time)

    where I show a method without using vector equation of lines
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    Thanks so much for the help! Yes the answer in the back of the book was k=2

    I get how to do it now, your help was useful! I eventually told my teacher in lesson that I hadn't done it because I couldn't, and he said he wasn't surprised to be honest because he hadn't taught us how to do the equation of a line yet with vectors, he wanted to see if any of us could work it out, or at least have a go at something still haven't covered equation of a line actually, apparently that's next lesson *sigh*
    Thanks for the help though
 
 
 
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