So managed to do part i), how do you do part ii)?
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h26
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 19022018 12:42

the bear
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 19022018 12:46
this is MEI yeah ?

h26
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 19022018 12:47
(Original post by the bear)
this is MEI yeah ? 
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 19022018 12:48
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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 19022018 12:54
(Original post by h26)
So managed to do part i), how do you do part ii)? 
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 19022018 15:22
(Original post by RDKGames)
Vector GF is the projection of the vector GH onto the xyplane. That vector can be written as 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.
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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 19022018 15:26
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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 19022018 15:27
(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.
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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 19022018 15:30
(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 
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 19022018 15:40
(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.
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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 19022018 15:44
(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
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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 19022018 15:48
(Original post by the bear)
the third coordinate is the "up" direction which is not north on a map ?
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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 19022018 16:18
(Original post by RDKGames)
Vector GF is the projection of the vector GH onto the xyplane. That vector can be written as 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.
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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 19022018 16:27
(Original post by h26)
Thanks very much But vector GF is not (1+t,0.5+2t,0), it is (t, 2t,0).
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?Last edited by RDKGames; 19022018 at 16:29. 
h26
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 19022018 16:55
(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 " is the angle that the path of motion makes with the ground"
No. Read the question, it says that represents north. 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.
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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 19022018 17:13
(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?
Also, how is theta the angle that the path makes with the ground? 
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 19022018 17:19
(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. 
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 19022018 17:25
(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? 
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 19022018 17:26
(Original post by RDKGames)
Yes GF is parallel to neither x nor y, so it's made up from the x and y components.
"Please rate some other members before rating this member again."  Lol this is what happened when I tried to give a repLast edited by h26; 19022018 at 17:28. Reason: Extra added 
the bear
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 19022018 21:49
(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
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