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    A tunnel is to be excavated through a hill. In order to define position, co-ordinates (x,y,z) are taken relative to an origin O such that x is the distance east from O, y is the distance north and z is the vertical distance upwards, with one unit equal to 100m.

    The tunnel starts at point A(2,3,5) and runs in direction (1,1,-0.5). It meets the hillside again at B.

    Right first of all vertical distance upwards? Isn't the same as north ie. y?

    Find the angle which AB makes with the upward vertical.

    Don't work this out! I just have a question, they used the upward vertical movement as (0,0,1) but why 1? And if you use any other value, why don't you get the same angle if AB is meeting the plane, the angle should be same as at (0,0,2) ?? Sorry confused!
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    "vertical distance upwards" = "height above O".

    The upward vertical has vector direction (0, 0, 1). The required angle theta satisfies

    cos(theta) = [(1, 1, -0.5) . (0, 0, 1)] / (length(1, 1, -0.5) * length(0, 0, 1)).

    You can take the upward vertical direction to be (0, 0, 2004) if you prefer! You will get the same value of cos(theta) because the numerator and denominator will both be multiplied by 2004.

    Since the tunnel slopes down from A, theta should be between 90 and 180 degrees.
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    (Original post by Jonny W)
    "vertical distance upwards" = "height above O".

    The upward vertical has vector direction (0, 0, 1). The required angle theta satisfies

    cos(theta) = [(1, 1, -0.5) . (0, 0, 1)] / (length(1, 1, -0.5) * length(0, 0, 1)).

    You can take the upward vertical direction to be (0, 0, 2004) if you prefer! You will get the same value of cos(theta) because the numerator and denominator will both be multiplied by 2004.

    Since the tunnel slopes down from A, theta should be between 90 and 180 degrees.
    Hm okies so if they say a point P on AB is directly above a point Q in something else, this means z will be different but x and y will be the same?
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    (Original post by Sugaray)
    Hm okies so if they say a point P on AB is directly above a point Q in something else, this means z will be different but x and y will be the same?
    Yes.
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    Okies, well if there is a vector perpendicular to 2 of the translations of a plane, does that mean something? The question was:

    The position of 3 points A,B, C on a plane ski-slope are
    a=4i + 2j - k
    b= -2i + 26j
    c=16i + 17j +2k
    where units are metres.

    Show that the vector 2i - 3j + 7k is perpendicular to A->B and also perpendicular to A->C . Hence find the equation of the plane of the ski-slope.

    Ok I've done the perpendicular bits and I'm doing the equation of the plane the long way by solving all the equations but i'm getting the feeling that the 'hence' bit means you can get it straight away by knowing this vector is perpendicular? Because the equation of the plane is apparently 2x - 3y +7z +5 =0 so can I get this straight from the vector somehow with the 5 coming up somewhere?

    Thank you Thank you Thank you Thank you Thank you
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    (r - s).n = 0 is the equation of the plane that
    - includes the point s,
    - has normal vector n.

    This is because saying that (r - s).n = 0 is the same as saying that (r - s) is perpendicular to n.

    In your case,

    n = 2i - 3j + 7k

    and you can take s to be a, b or c (it doesn't matter which - you'll get the same answer).
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    Thank you
 
 
 
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Updated: June 17, 2004
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