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    So managed to do part i), how do you do part ii)?
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    this is MEI yeah ?
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    (Original post by the bear)
    this is MEI yeah ?
    Yep
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    ii) put in two values of t into the position vector formula. then use the coordinates to work out the angle you need.
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    (Original post by h26)
    So managed to do part i), how do you do part ii)?
    Vector GF is the projection of the vector GH onto the xy-plane. That vector can be written as \overrightarrow{GF}=(1+t, 0.5+2t,0) which means that the line GF had direction vector (1,2,0). Now just find the angle between GF and GH using their direction vectors.
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    (Original post by RDKGames)
    Vector GF is the projection of the vector GH onto the xy-plane. That vector can be written as \overrightarrow{GF}=(1+t, 0.5+2t,0) which means that the line GF had direction vector (1,2,0). Now just find the angle between GF and GH using their direction vectors.
    Thanks very much- But vector GF is not (1+t,0.5+2t,0), it is (t, 2t,0). I got this using the fact that the position vector of G is (1,0.5,0).

    Furthermore, the position vector of H keeps changing.The position vector of H is given by: (1+t,0.5+2t, 2t)
    We can rewrite this:
    (1+t,0.5+2t, 2t) = (1, 0.5, 0) + (t, 2t, 2t) where (1,0.5,0) is the position vector of G and (t,2t, 2t) is the direction vector GH.

    Also: |GH|= √(t^2 + 4t^2 +4t^2) = 3t so the magnitude of GH is 3t

    Anyway, back to the question..
    To work out the magnitude of GF "|GF|", you do √((t)^2 +(2t)^2) = √(5t^2) = t√5

    Since |FH|= 2t and |GF|= t√5 , you can find theta by doing tantheta =(2t)/t√5 = 2/√5 so theta =tan^-1(2/√5) =41.8 degrees which is the answer

    #I think they've worded the question poorly as theta is not really the angle the flight path makes with the the horizontal.

    I am confused a bit with the diagram they have drawn for part iii) in the mark scheme: I thought north is the z axis?

    I've attached the mark scheme below for part i),ii) and iii):
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    Could you kindly let me know your thoughts.Many thanks
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    as well as zero which yields ( 1, 0.5. 0 )you could put in t = 1 to get ( 2, 2.5, 2) ; these two points will give you the angle.
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    (Original post by the bear)
    ii) put in two values of t into the position vector formula. then use the coordinates to work out the angle you need.
    Thanks very much- my post to RDKGames shows how I got there in the end with ii) but I am confused with part iii) now..
    I am confused a bit with the diagram they have drawn for part iii) in the mark scheme: I thought north is the z axis?

    I've attached the mark scheme below for part i),ii) and iii):

    Could you kindly let me know your thoughts.Many thanks
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    (Original post by h26)
    Thanks very much- my post to RDKGames shows how I got there in the end with ii) but I am confused with part iii) now..
    I am confused a bit with the diagram they have drawn for part iii) in the mark scheme: I thought north is the z axis?

    I've attached the mark scheme below for part i),ii) and iii):

    Could you kindly let me know your thoughts.Many thanks
    this just uses the x and y coordinates ( 1, 0.5 )and ( 2, 2.5) to find the angle clockwise from the y direction.
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    (Original post by the bear)
    this just uses the x and y coordinates ( 1, 0.5 )and ( 2, 2.5) to find the angle clockwise from the y direction.
    Thanks very much- but I don't quite understand the method you are using here ..so G is (1,0.5) and F is (2,2.5)? I am not really sure:confused:

    Also,(if we go with the mark scheme method) since north is the z axis, then shouldn't we be finding the angle clockwise from the z direction?
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    (Original post by h26)
    Thanks very much- but I don't quite understand the method you are using here ..so G is (1,0.5) and F is (2,2.5)? I am not really sure:confused:

    Also,(if we go with the mark scheme method) since north is the z axis, then shouldn't we be finding the angle clockwise from the z direction?
    the third coordinate is the "up" direction which is not north on a map ?
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    (Original post by the bear)
    the third coordinate is the "up" direction which is not north on a map ?
    Thanks but what do you mean?
    The third coordinate is the z coordinate, so it is the "up" direction coordinate. The z axis represents north, so it is north on the map, right?
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    (Original post by RDKGames)
    Vector GF is the projection of the vector GH onto the xy-plane. That vector can be written as \overrightarrow{GF}=(1+t, 0.5+2t,0) which means that the line GF had direction vector (1,2,0). Now just find the angle between GF and GH using their direction vectors.
    Thanks very much- But vector GF is not (1+t,0.5+2t,0), it is (t, 2t,0). I got this using the fact that the position vector of G is (1,0.5,0).

    Furthermore, the position vector of H keeps changing.The position vector of H is given by: (1+t,0.5+2t, 2t)
    We can rewrite this:
    (1+t,0.5+2t, 2t) = (1, 0.5, 0) + (t, 2t, 2t) where (1,0.5,0) is the position vector of G and (t,2t, 2t) is the direction vector GH.

    Also: |GH|= √(t^2 + 4t^2 +4t^2) = 3t so the magnitude of GH is 3t

    Anyway, back to the question..
    To work out the magnitude of GF "|GF|", you do √((t)^2 +(2t)^2) = √(5t^2) = t√5

    Since |FH|= 2t and |GF|= t√5 , you can find theta by doing tantheta =(2t)/t√5 = 2/√5 so theta =tan^-1(2/√5) =41.8 degrees which is the answer

    #I think they've worded the question poorly as theta is not really the angle the flight path makes with the the horizontal.

    I am confused a bit with the diagram they have drawn for part iii) in the mark scheme: I thought north is the z axis?

    I've attached the mark scheme below for part i),ii) and iii):


    Could you kindly let me know your thoughts.Many thanks
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    (Original post by h26)
    Thanks very much- But vector GF is not (1+t,0.5+2t,0), it is (t, 2t,0).
    Yes, you're right. I was is a lecture when I wrote that so I must've zoomed out

    Anyway, back to the question..
    To work out the magnitude of GF "|GF|", you do √((t)^2 +(2t)^2) = √(5t^2) = t√5

    Since |FH|= 2t and |GF|= t√5 , you can find theta by doing tantheta =(2t)/t√5 = 2/√5 so theta =tan^-1(2/√5) =41.8 degrees which is the answer
    Yes. The alternative is to consider the dot product of vectors GF and GH which is what I was leaning towards.

    I think they've worded the question poorly as theta is not really the angle the flight path makes with the the horizontal.
    Yes, the correct wording would be "\theta is the angle that the path of motion makes with the ground"

    I am confused a bit with the diagram they have drawn for part iii) in the mark scheme: I thought north is the z axis?
    No. Read the question, it says that y represents north. z is just the elevation coordinate. Do you understand their working? They are considering the xy plane where y is north and x is east. All they are doing is working with the projection of the motion vector onto the xy plane, which is GF, and finding its bearing from the y axis since bearings are measured from the north.
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    (Original post by RDKGames)
    Yes, you're right. I was is a lecture when I wrote that so I must've zoomed out



    Yes. The alternative is to consider the dot product of vectors GF and GH which is what I was leaning towards.



    Yes, the correct wording would be "\theta is the angle that the path of motion makes with the ground"



    No. Read the question, it says that y represents north. z is just the elevation coordinate. Do you understand their working? They are considering the xy plane where y is north and x is east. All they are doing is working with the projection of the motion vector onto the xy plane, which is GF, and finding its bearing from the y axis since bearings are measured from the north.
    Haha no worries Thanks a lot for the help - the dot product way was so much easier!

    So am I right in saying z axis represents up and down(the plane), y axis represents north and south, x axis represents east and west in all situations? Or do you get to choose your north direction?

    Also, how is theta the angle that the path makes with the ground?
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    (Original post by h26)
    So am I right in saying z axis represents up and down(the plane), y axis represents north and south, x axis represents east and west in all situations? Or do you get to choose your north direction?
    In this question, it is given to you so you avoid ambiguity. Though if it wasn't given to you, then you may as well 'assume' that y is the north/south, x is the east/west, and z is the up/down as this is the common notion.

    Also, how is theta the angle that the path makes with the ground?
    Because the xy plane is the ground and the angle shown is clearly the angle between the vector of motion and the xy plane.
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    (Original post by RDKGames)
    In this question, it is given to you so you avoid ambiguity. Though if it wasn't given to you, then you may as well 'assume' that y is the north/south, x is the east/west, and z is the up/down as this is the common notion.



    Because the xy plane is the ground and the angle shown is clearly the angle between the vector of motion and the xy plane.
    Ohh that makes sense! And vector GF is in the xy plane too however it isn't heading in the x or y direction cause it's made up of some units in the y direction and some units in the x direction, right?
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    (Original post by h26)
    Ohh that makes sense! And vector GF is in the xy plane too however it isn't heading in the x or y direction cause it's made up of some units in the y direction and some units in the x direction, right?
    Yes GF is parallel to neither x nor y, so it's made up from the x and y components.
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    (Original post by RDKGames)
    Yes GF is parallel to neither x nor y, so it's made up from the x and y components.
    Thanks a lot!!! You really helped to clear things!

    "Please rate some other members before rating this member again." - Lol this is what happened when I tried to give a rep
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    (Original post by h26)
    Thanks a lot!!! You really helped to clear things!

    "Please rate some other members before rating this member again." - Lol this is what happened when I tried to give a rep
    it's OK... RDK has enough rep already
 
 
 
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