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A vector a has magnitude |a| = 7 and direction θ = −70°. Calculate the component form of a, giving the components correct to two decimal places.
Applying the equation a = |a| cos θi + |a| sin θj, where |a| = 7 and θ = −70°, we obtain the component form
Find the magnitude and direction of each of the vectors a = −3i + 2j, b = 6i − j and of their sum a + b, giving your answers correct to one decimal place.
In each case, the strategy on page 22 may be applied. The magnitude of the vector a = −3i + 2j is
Since the components are a1 = −3, a2 = 2, we have
Also, (−3, 2) lies in the second quadrant, so the direction of a is θ = 180° − φ ≈ 146.3°.
The magnitude of the vector b = 6i − j is
Since the components are b1 = 6, b2 = −1, we have
Also, (6, −1) lies in the fourth quadrant, so the direction of b is θ = − φ ≈ −9.5°.
The sum of the vectors a and b is
The magnitude of this vector is
Since the components are c1 = 3, c2 = 1, we have
Also, (3, 1) lies in the first quadrant, so the direction of c is θ = φ ≈ 18.4°.
The displacement from Derby to Birmingham is 57 km at S 30° W. The displacement from Derby to Leicester is 32 km at S 45° E. In terms of direct distance and a bearing, find:
the displacement from Leicester to Derby;
the displacement from Leicester to Birmingham, giving your answers correct to one decimal place.
(a) The displacement from Leicester to Derby is the opposite of the given displacement from Derby to Leicester, so it is 32 km at N 45°W.
(b) Let i be 1 km East and let j be 1 km North. Denote the displacement from Leicester to Derby by the vector a, and the displacement from Derby to Birmingham by the vector b. Then the required displacement from Leicester to Birmingham is represented by the vector a + b. These vectors are sketched in Figure S.21.
Figure S.21
Long description
The vector a has magnitude |a| = 32 and direction 90° + 45° = 135°, so its component form is
The vector b has magnitude |b| = 57 and direction −(90° + 30°) = −120°, so its component form is
The resultant is
The strategy on page 22 may now be applied. The vector c has magnitude
Since the components are c1 ≈ −51.13, c2 ≈ −26.73, we have
Also, (−51.13, −26.73) lies in the third quadrant, so the direction of c is θ = −(180° −φ) ≈ −152.4°. This corresponds to the bearing S 62.4°W.
Hence the displacement from Leicester to Birmingham is 57.7 km at S 62.4°W.
An aeroplane has a speed in still air of 50 ms−1 and is pointed in the direction N 20° E, but it flies in a wind of speed 10 ms−1 blowing from N 70° W. Find the velocity of the aeroplane relative to the ground, in terms of its speed and a bearing. Give your answers correct to one decimal place.
Let i be 1 m s−1 East and let j be 1 ms−1 North. Suppose that va is the velocity of the aeroplane in still air and vw is the velocity of the wind. These vectors are shown in Figure S.22.
Figure S.22
Long description
From the information given, va has magnitude |va| = 50 and direction θa = 70°.
The wind comes from N 70°W and hence blows towards S 70°E, for which the direction is −(90° −70°). Hence the vector vw has magnitude |vw| = 10 and direction θw = −20°.
The component forms of the vectors are therefore
The resultant velocity is
The strategy on page 22 may now be applied. The resultant speed of the aeroplane is
Since the components of v are v1 ≈ 26.50, v2 ≈ 43.56, we have
Also, (26.50, 43.56) lies in the first quadrant, so the direction of v is θ = φ ≈ 58.7°. This corresponds to the bearing N 31.3°E.
Thus the velocity of the aeroplane relative to the ground is 51.0 m s−1 at N 31.3°E.
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