# Maths vectors question

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

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===

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===

0

reply

Report

#2

(Original post by

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.

**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.

0

reply

I tried the cosine rule but I didnt get very far because the vector is in terms of t, and they dont cancel out

Posted from TSR Mobile

Posted from TSR Mobile

0

reply

Report

#4

(Original post by

I tried the cosine rule but I didnt get very far because the vector is in terms of t, and they dont cancel out

Posted from TSR Mobile

**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

Posted from TSR Mobile

Then

EDIT: You can also do it all in terms of but it must obviously cancel out.

0

reply

Report

#5

(Original post by

I tried the cosine rule but I didnt get very far because the vector is in terms of t, and they dont cancel out

**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

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?

0

reply

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)

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)

0

reply

Report

#9

(Original post by

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)

**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 simply need to find and by Pythagoras' Theorem, then by trigonometry (and aid of the diagram) you know that

Just to add on, your and are incorrect

0

reply

Report

#11

Coordinate G(1,0.5,0)

Vector GH = i+2j+2k

So coordinate F(2,2.5,0)

Vector GF =i+2j

Magnitude GF =square root 5

Magnitude GH=3

Scalar Product GF x GH=5

Cos x=Scalar Product/Magnitude GH x Magnitude GF

So x,the angle of the flight is 41.8 to 1 decimal place

Vector GH = i+2j+2k

So coordinate F(2,2.5,0)

Vector GF =i+2j

Magnitude GF =square root 5

Magnitude GH=3

Scalar Product GF x GH=5

Cos x=Scalar Product/Magnitude GH x Magnitude GF

So x,the angle of the flight is 41.8 to 1 decimal place

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top