avsl020
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The flight path of a helicopter H taking off from an airport.
Coordinate axes Oxyz are set up with the origin O at the base of the airport control tower. The x-
axis is due east, the y-axis due north, and the z-axis vertical. The units of distance are kilometres
throughout.
The helicopter takes off from the point G. The position vector r of the helicopter t minutes after
take-off is given by: r=(1+t)i+(0.5+2t)j+2tk .

(i) Write down the coordinates of G - (1,0.5.0)
(ii) Find the angle the flight path makes with the horizontal.

===Moved to Maths===
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Kevin De Bruyne
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(Original post by avsl020)
The flight path of a helicopter H taking off from an airport.
Coordinate axes Oxyz are set up with the origin O at the base of the airport control tower. The x-
axis is due east, the y-axis due north, and the z-axis vertical. The units of distance are kilometres
throughout.
The helicopter takes off from the point G. The position vector r of the helicopter t minutes after
take-off is given by: r=(1+t)i+(0.5+2t)j+2tk .

(i) Write down the coordinates of G - (1,0.5.0)
(ii) Find the angle the flight path makes with the horizontal.
What have you tried so far? A diagram may help, and to draw a triangle on that diagram.
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avsl020
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I tried the cosine rule but I didnt get very far because the vector is in terms of t, and they dont cancel out

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RDKGames
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(Original post by avsl020)
I tried the cosine rule but I didnt get very far because the vector is in terms of t, and they dont cancel out

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It's quite simple if you just take the position of H to be at an arbitrary time such as t=1, and then you can deduce the coordinates for F.

Then \vert {\bf GH} \vert \cos(\theta) = \vert {\bf GF} \vert

EDIT: You can also do it all in terms of t but it must obviously cancel out.
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DFranklin
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(Original post by avsl020)
I tried the cosine rule but I didnt get very far because the vector is in terms of t, and they dont cancel out
Well, can you find a vector that is in the same direction as the flight path? (You should actually be able to write one down without any work)!

Call that vector v and split out the horizontal and vertical components. (i.e. the part in the xy plane and the part that's vertical).

What's the size of the horizontal component?
What's the size of the vertical component?

So the angle the vector makes with the horizontal is going to be?
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avsl020
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So coordinates of F: (0, 2.5, 0) ?

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RDKGames
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(Original post by avsl020)
So coordinates of F: (0, 2.5, 0) ?

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The coordinates of F would be the same as H except the z-coordinate is 0.
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avsl020
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Ok, so when t=1, the vector G to H is 2i+2.5j+2k,
Coordinates of H: (2, 2.5, 2)
Coordinates of F: (2, 2.5, 0)

scalar product= (2×2)+(2.5*2.5)= 10.25

|GH|= square root of (4+6.25+4) = root 54/2

|GF|= square root of (4+6.25) = root 10.23

When i sub everything into the equation and then do the inverse to work out theta, i get 31.99 degrees which is the wrong answer (it's 41.8)
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RDKGames
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(Original post by avsl020)
Ok, so when t=1, the vector G to H is 2i+2.5j+2k,
Coordinates of H: (2, 2.5, 2)
Coordinates of F: (2, 2.5, 0)

scalar product= (2×2)+(2.5*2.5)= 10.25

|GH|= square root of (4+6.25+4) = root 54/2

|GF|= square root of (4+6.25) = root 10.23

When i sub everything into the equation and then do the inverse to work out theta, i get 31.99 degrees which is the wrong answer (it's 41.8)
You don't really need to carry out any scalar products.

You simply need to find \vert {\bf GH } \vert and \vert {\bf GF } \vert by Pythagoras' Theorem, then by trigonometry (and aid of the diagram) you know that \vert {\bf GH} \vert \cos(\theta) = \vert {\bf GF} \vert

Just to add on, your \vert {\bf GH } \vert and \vert {\bf GF } \vert are incorrect
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