Vectors Introduction & 2D

2 Two forces \(\mathbf { F } _ { 1 } \mathrm {~N}\) and \(\mathbf { F } _ { 2 } \mathrm {~N}\) are given by \(\mathbf { F } _ { 1 } = - 6 \mathbf { i } + 2 \mathbf { j }\) and \(\mathbf { F } _ { 2 } = - 8 \mathbf { i } + \mathbf { j }\).
Show that the magnitude of the resultant of these two forces is \(\sqrt { 205 } \mathrm {~N}\).

\includegraphics{figure_6} The diagram shows two horizontal forces \(\mathbf{P}\) and \(\mathbf{Q}\) acting at the origin \(O\) of rectangular coordinates \(Oxy\). The components of \(\mathbf{P}\) in the \(x\)- and \(y\)-directions are 12 N and 17 N respectively. The components of \(\mathbf{Q}\) in the \(x\)- and \(y\)-directions are -5 N and 7 N respectively.

1. Write down the components, in the \(x\)- and \(y\)-directions, of the resultant of \(\mathbf{P}\) and \(\mathbf{Q}\). [2]
2. Hence, or otherwise, calculate the magnitude of this resultant and the angle the resultant makes with the positive \(x\)-axis. [4]

4 \includegraphics[max width=\textwidth, alt={}, center]{2bb3c9bb-60f0-440d-a148-b4db3478ca31-2_387_1091_2019_525} Three coplanar forces of magnitudes \(8 \mathrm {~N} , 12 \mathrm {~N}\) and 2 N act at a point. The resultant of the forces has magnitude \(R \mathrm {~N}\). The directions of the three forces and the resultant are shown in the diagram. Find \(R\) and \(\theta\).

Two ships \(A\) and \(B\) are sailing in the same direction at constant speeds of 12 km h\(^{-1}\) and 16 km h\(^{-1}\) respectively. They are sailing along parallel lines which are 4 km apart. When the distance between the ships is 4 km, \(B\) turns through 30° towards \(A\). Find the shortest distance between the ships in the subsequent motion. [7]

[In this question \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal unit vectors due east and due north respectively.] A hiker \(H\) is walking with constant velocity \((1.2\mathbf{i} - 0.9\mathbf{j})\) m s\(^{-1}\).

1. Find the speed of \(H\). [2]

\includegraphics{figure_3} A horizontal field \(OABC\) is rectangular with \(OA\) due east and \(OC\) due north, as shown in Figure 3. At twelve noon hiker \(H\) is at the point \(Y\) with position vector \(100\mathbf{j}\) m, relative to the fixed origin \(O\).

1. Write down the position vector of \(H\) at time \(t\) seconds after noon. [2]

At noon, another hiker \(K\) is at the point with position vector \((9\mathbf{i} + 46\mathbf{j})\) m. Hiker \(K\) is moving with constant velocity \((0.75\mathbf{i} + 1.8\mathbf{j})\) m s\(^{-1}\).

1. Show that, at time \(t\) seconds after noon, $$\overrightarrow{HK} = [(9 - 0.45t)\mathbf{i} + (2.7t - 54)\mathbf{j}] \text{ metres.}$$ [4]

Hence,

1. show that the two hikers meet and find the position vector of the point where they meet. [5]

\includegraphics{figure_6} A ship \(P\) is moving with constant velocity 7 m s\(^{-1}\) in the direction with bearing 110°. A second ship \(Q\) is moving with constant speed 10 m s\(^{-1}\) in a straight line. At one instant \(Q\) is at the point \(X\), and \(P\) is 7400 m from \(Q\) on a bearing of 050° (see diagram). In the subsequent motion, the shortest distance between \(P\) and \(Q\) is 1790 m.

1. Show that one possible direction for the velocity of \(Q\) relative to \(P\) has bearing 036°, to the nearest degree, and find the bearing of the other possible direction of this relative velocity. [3]

Given that the velocity of \(Q\) relative to \(P\) has bearing 036°, find

1. the bearing of the direction in which \(Q\) is moving, [4]
2. the magnitude of the velocity of \(Q\) relative to \(P\), [2]
3. the time taken for \(Q\) to travel from \(X\) to the position where the two ships are closest together, [3]
4. the bearing of \(P\) from \(Q\) when the two ships are closest together. [1]

\includegraphics{figure_1}

1. \(A\) has velocity \(\vec{x}\) and \(C\) has velocity \(\vec{v}\)

\(A\) has velocity \(\vec{v}\), there are velocities \(\vec{x}\), \(\vec{v}\), \(\vec{v}\) around point \(O\), and velocity \(\vec{v}\)

1. 1. —
   2. \(v\) and \(v\)

6. Two particles \(P\) and \(Q\) have constant velocity vectors \(\mathbf { v } _ { P }\) and \(\mathbf { v } _ { Q }\) respectively. The magnitude of the velocity of \(P\) relative to \(Q\) is equal to the speed of \(P\). If the direction of motion of one of the particles is reversed, the magnitude of the velocity of \(P\) relative to \(Q\) is doubled. Find

1. the ratio of the speeds of \(P\) and \(Q\),
2. the cosine of the angle between the directions of motion of \(P\) and \(Q\).

1. The points \(P , Q\) and \(R\) have coordinates (-3, 7), (9, 11) and (12, 2) respectively.
   1. Prove that angle \(P Q R = 90 ^ { \circ }\)
   Given that the point \(S\) is such that \(P Q R S\) forms a rectangle,
2. find the coordinates of \(S\).

1. The points \(P , Q\) and \(R\) have coordinates (-3, 7), (9, 11) and (12, 2) respectively.
   1. Prove that angle \(P Q R = 90 ^ { \circ }\)
   Given that the point \(S\) is such that \(P Q R S\) forms a rectangle,
2. find the coordinates of \(S\).

3. \begin{figure}[h] \end{figure} Figure 1 shows a regular pentagon \(O A B C D\). The vectors \(\mathbf { p }\) and \(\mathbf { q }\) are defined by \(\mathbf { p } = \overrightarrow { O A }\) and \(\mathbf { q } = \overrightarrow { O D }\) respectively. Let \(k\) be the number such that \(\overrightarrow { D B } = k \overrightarrow { O A }\).

1. Write down \(\overrightarrow { A C }\) in terms of \(\mathbf { p } , \mathbf { q }\) and \(k\) as appropriate.
2. Show that \(\overrightarrow { C D } = - \mathbf { p } - \frac { 1 } { k } \mathbf { q }\)
3. Hence find the value of \(k\) By considering triangle \(D B C\), or otherwise,
4. find the exact value of \(\sin 54 ^ { \circ }\)

A particle, \(P\), is moving with constant velocity \(8\mathbf{i} - 12\mathbf{j}\) A second particle, \(Q\), is moving with constant velocity \(a\mathbf{i} + 9\mathbf{j}\) \(Q\) travels in a direction which is parallel to the motion of \(P\). Find \(a\). Circle your answer. \(-6\) \quad \(-5\) \quad \(5\) \quad \(6\) [1 mark]

A particle, \(P\), is moving with constant velocity \(8\mathbf{i} - 12\mathbf{j}\) A second particle, \(Q\), is moving with constant velocity \(a\mathbf{i} + 9\mathbf{j}\) \(Q\) travels in a direction which is parallel to the motion of \(P\). Find \(a\). Circle your answer. \(-6\) \quad \(-5\) \quad \(5\) \quad \(6\) [1 mark]

4. In this question, \(\mathbf { i }\) and \(\mathbf { j }\) are perpendicular horizontal unit vectors and \(O\) is a fixed origin. A pedestrian moves with constant velocity \(\left[ \left( 2 q ^ { 2 } - 3 \right) \mathbf { i } + ( q + 2 ) \mathbf { j } \right] \mathrm { ms } ^ { - 1 }\). Given that the velocity of the pedestrian is parallel to the vector \(( \mathbf { i } - \mathbf { j } )\),

1. Show that one possible value of \(q\) is \({ } ^ { - } 1\) and find the other possible value of \(q\). Given that \(q = { } ^ { - } 1\), and that the pedestrian started walking at the point with position vector \(( 6 \mathbf { i } - \mathbf { j } ) \mathrm { m }\),
2. find the length of time for which the pedestrian is less than 5 m from \(O\).

1 The map of a large area of open land is marked in 1 km squares and a point near the middle of the area is defined to be the origin. The vectors \(\binom { 1 } { 0 }\) and \(\binom { 0 } { 1 }\) are in the directions east and north. At time \(t\) hours the position vectors of two hikers, Ashok and Kumar, are given by: $$\begin{array} { l l } \text { Ashok } & \mathbf { r } _ { \mathrm { A } } = \binom { - 2 } { 0 } + \binom { 8 } { 1 } t , \\ \text { Kumar } & \mathbf { r } _ { \mathrm { K } } = \binom { 7 t } { 10 - 4 t } . \end{array}$$

1. Prove that the two hikers meet and give the coordinates of the point where this happens.
2. Compare the speeds of the two hikers.

[In this question \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal unit vectors due east and due north respectively and position vectors are given relative to a fixed origin.] At 2 pm, the position vector of ship \(P\) is \((5\mathbf{i} - 3\mathbf{j})\) km and the position vector of ship \(Q\) is \((7\mathbf{i} + 5\mathbf{j})\) km.

1. Find the distance between \(P\) and \(Q\) at 2 pm. [3]

Ship \(P\) is moving with constant velocity \((2\mathbf{i} + 5\mathbf{j})\) km h\(^{-1}\) and ship \(Q\) is moving with constant velocity \((-3\mathbf{i} - 15\mathbf{j})\) km h\(^{-1}\).

1. Find the position vector of \(P\) at time \(t\) hours after 2 pm. [2]
2. Find the position vector of \(Q\) at time \(t\) hours after 2 pm. [1]
3. Show that \(Q\) will meet \(P\) and find the time at which they meet. [5]
4. Find the position vector of the point at which they meet. [2]

7. [In this question, the horizontal unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are directed due east and north respectively] A mountain rescue post \(O\) receives a distress call via a mobile phone from a walker who has broken a leg and cannot move. The walker says he is by a pipeline and he can also see a radio mast which he believes to be south-west of him. The pipeline is known to run north-south for a long distance through the point with position vector \(6 \mathbf { i } \mathrm {~km}\), relative to \(O\). The radio mast is known to be at the point with position vector \(2 \mathbf { j } \mathrm {~km}\), relative to \(O\).

1. Using the information supplied by the walker, write down his position vector in the form \(( a \mathbf { i } + b \mathbf { j } )\). The rescue party moves at a horizontal speed of \(5 \mathrm {~km} \mathrm {~h} ^ { - 1 }\). The leader of the party wants to give the walker and idea of how long it will take to for the rescue party to arrive.
2. Calculate how long it will take for the rescue party to reach the walker's estimated position. The rescue party sets out and walks straight towards the walker's estimated position at a constant horizontal speed of \(5 \mathrm {~km} \mathrm {~h} ^ { - 1 }\). After the party has travelled for one hour, the walker rings again. He is very apologetic and says that he now realises that the radio mask is in fact north-west of his position
3. Find the position vector of the walker.
4. Find in degrees to one decimal place, the bearing on which the rescue party should now travel in order to reach the walker directly. \section*{END}

5 You are given that \(\overrightarrow { \mathrm { OA } } = \binom { 3 } { - 1 }\) and \(\overrightarrow { \mathrm { OB } } = \binom { 5 } { - 3 }\). Determine the exact length of \(A B\).

One of the following is an expression for the distance between the points represented by position vectors \(5\mathbf{i} - 3\mathbf{j}\) and \(18\mathbf{i} + 7\mathbf{j}\) Identify the correct expression. Tick (\(\checkmark\)) one box. [1 mark] \(\sqrt{13^2 + 4^2}\) \(\sqrt{13^2 + 10^2}\) \(\sqrt{23^2 + 4^2}\) \(\sqrt{23^2 + 10^2}\)

5 Points \(A , B , C\) and \(D\) have position vectors \(\mathbf { a } = \binom { 1 } { 2 } , \mathbf { b } = \binom { 3 } { 5 } , \mathbf { c } = \binom { 7 } { 4 }\) and \(\mathbf { d } = \binom { 4 } { k }\).

1. Find the value of \(k\) for which \(D\) is the midpoint of \(A C\).
2. Find the two values of \(k\) for which \(| \overrightarrow { A D } | = \sqrt { 13 }\).
3. Find one value of \(k\) for which the four points form a trapezium.

5.
[In this question the unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are horizontal vectors due east and due north respectively and position vectors are given relative to a fixed origin.]

1. A particle \(P\) is moving with constant velocity. The position vector of \(P\) at time \(t\) seconds \(( t \geqslant 0 )\) is \(\mathbf { r }\) metres, relative to a fixed origin \(O\), and is given by

$$\mathbf { r } = ( 2 t - 3 ) \mathbf { i } + ( 4 - 5 t ) \mathbf { j }$$

1. Find the initial position vector of \(P\). The particle \(P\) passes through the point with position vector \(( 3.4 \mathbf { i } - 12 \mathbf { j } )\) m at time \(T\) seconds.
2. Find the value of \(T\).
3. Find the speed of \(P\).

8. [In this question \(\mathbf { i }\) and \(\mathbf { j }\) are horizontal unit vectors directed due east and due north respectively and position vectors are given relative to a fixed origin.] At 7 am a ship leaves a port and moves with constant velocity. The position vector of the port is \(( - 2 \mathbf { i } + 9 \mathbf { j } ) \mathrm { km }\). At 7.36 am the ship is at the point with position vector \(( 4 \mathbf { i } + 6 \mathbf { j } ) \mathrm { km }\).

1. Show that the velocity of the ship is \(( 10 \mathbf { i } - 5 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }\)
2. Find the position vector of the ship \(t\) hours after leaving port. At 8.48 am a passenger on the ship notices that a lighthouse is due east of the ship. At 9 am the same passenger notices that the lighthouse is now north east of the ship.
3. Find the position vector of the lighthouse.
4. Find the position vector of the ship when it is due south of the lighthouse.
   
   \includegraphics[max width=\textwidth, alt={}]{151d9232-5a78-4bc1-a57e-6c9cae80e473-32_2258_53_308_1980}

9 Two forces \(( 3 \mathbf { i } + 2 \mathbf { j } ) \mathrm { N }\) and \(\mathbf { F N }\) act on a particle \(P\) of mass 4 kg .
Given that the acceleration of \(P\) is \(( - 2 \mathbf { i } + 3 \mathbf { j } ) \mathrm { ms } ^ { - 2 }\), calculate \(\mathbf { F }\).

3. A particle \(P\) of mass 0.4 kg moves under the action of a single constant force \(\mathbf { F }\) newtons. The acceleration of \(P\) is \(( 6 \mathbf { i } + 8 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 2 }\). Find

1. the angle between the acceleration and \(\mathbf { i }\),
2. the magnitude of \(\mathbf { F }\). At time \(t\) seconds the velocity of \(P\) is \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\). Given that when \(t = 0 , \mathbf { v } = 9 \mathbf { i } - 10 \mathbf { j }\), (c) find the velocity of \(P\) when \(t = 5\).

2. \begin{figure}[h] \end{figure} Figure 1 shows a regular hexagon \(O P Q R S T\).
The vectors \(\mathbf { p }\) and \(\mathbf { q }\) are defined by \(\mathbf { p } = \overrightarrow { O P }\) and \(\mathbf { q } = \overrightarrow { O Q }\) Forces, in Newtons, \(\mathbf { F } _ { P } = ( \overrightarrow { O P } ) , \mathbf { F } _ { Q } = 2 \times ( \overrightarrow { O Q } ) , \mathbf { F } _ { R } = 3 \times ( \overrightarrow { O R } ) , \mathbf { F } _ { S } = 4 \times ( \overrightarrow { O S } )\) and \(\mathbf { F } _ { T } = 5 \times ( \overrightarrow { O T } )\) are applied to a particle.

1. Find, in terms of \(\mathbf { p }\) and \(\mathbf { q }\), the resultant force on the particle. The magnitude of the acceleration of the particle due to these forces is \(13 \mathrm {~ms} ^ { - 2 }\) Given that the mass of the particle is 3 kg ,
2. find \(| \mathbf { p } |\) \includegraphics[max width=\textwidth, alt={}, center]{71cd126f-1c7d-4e37-a26d-7ff98a74fd79-04_2255_56_310_1980}

3 The points \(P , Q\) and \(R\) have coordinates \(( - 1,6 ) , ( 2,10 )\) and \(( 11,1 )\) respectively. Find the angle \(P R Q\).

4. \begin{figure}[h] \end{figure} Figure 2 shows the points \(A\) and \(B\) with position vectors \(\mathbf { a }\) and \(\mathbf { b }\) respectively, relative to a fixed origin \(O\). Given that \(| \mathbf { a } | = 5 , | \mathbf { b } | = 6\) and a.b \(= 20\)

1. find the cosine of angle \(A O B\),
2. find the exact length of \(A B\).
3. Show that the area of triangle \(O A B\) is \(5 \sqrt { 5 }\)

1. (i) Two non-zero vectors, \(\mathbf { a }\) and \(\mathbf { b }\), are such that

$$| \mathbf { a } + \mathbf { b } | = | \mathbf { a } | + | \mathbf { b } |$$ Explain, geometrically, the significance of this statement.
(ii) Two different vectors, \(\mathbf { m }\) and \(\mathbf { n }\), are such that \(| \mathbf { m } | = 3\) and \(| \mathbf { m } - \mathbf { n } | = 6\) The angle between vector \(\mathbf { m }\) and vector \(\mathbf { n }\) is \(30 ^ { \circ }\) Find the angle between vector \(\mathbf { m }\) and vector \(\mathbf { m } - \mathbf { n }\), giving your answer, in degrees, to one decimal place.

A cyclist \(P\) is cycling due north at a constant speed of 20 km h\(^{-1}\). At 12 noon another cyclist \(Q\) is due west of \(P\). The speed of \(Q\) is constant at 10 km h\(^{-1}\). Find the course which \(Q\) should set in order to pass as close to \(P\) as possible, giving your answer as a bearing. [5]

6 At noon, two ships, \(A\) and \(B\), are a distance of 12 km apart, with \(B\) on a bearing of \(065 ^ { \circ }\) from \(A\). The ship \(B\) travels due north at a constant speed of \(10 \mathrm {~km} \mathrm {~h} ^ { - 1 }\). The ship \(A\) travels at a constant speed of \(18 \mathrm {~km} \mathrm {~h} ^ { - 1 }\). \includegraphics[max width=\textwidth, alt={}, center]{a90a2de3-5cc0-4e87-b29a-2562f86eee17-16_492_585_445_738}

1. Find the direction in which \(A\) should travel in order to intercept \(B\). Give your answer as a bearing.
2. In fact, the ship \(A\) actually travels on a bearing of \(065 ^ { \circ }\).
   1. Find the distance between the ships when they are closest together.
   2. Find the time when the ships are closest together.

4 \includegraphics[max width=\textwidth, alt={}, center]{ea62d6d9-ac2f-44e7-8bfb-ae9aeea7109b-2_688_777_1382_683} From a boat \(B\), a cruiser \(C\) is observed 3500 m away on a bearing of \(040 ^ { \circ }\). The cruiser \(C\) is travelling with constant speed \(15 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) along a straight line course with bearing \(110 ^ { \circ }\) (see diagram). The boat \(B\) travels with constant speed \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) on a straight line course which takes it as close as possible to the cruiser \(C\).

1. Show that the bearing of the course of \(B\) is \(073 ^ { \circ }\), correct to the nearest degree.
2. Find the magnitude and the bearing of the velocity of \(C\) relative to \(B\).
3. Find the shortest distance between \(B\) and \(C\) in the subsequent motion.

9 Three forces \(\binom { 7 } { - 6 } \mathrm {~N} , \binom { 2 } { 5 } \mathrm {~N}\) and \(\mathbf { F N }\) act on a particle.
Given that the particle is in equilibrium under the action of these three forces, calculate \(\mathbf { F }\).

A particle \(P\) is acted upon by three forces \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_3\) given by \(\mathbf{F}_1 = (6\mathbf{i} - 4\mathbf{j}) \text{ N}\), \(\mathbf{F}_2 = (-3\mathbf{i} + 9\mathbf{j}) \text{ N}\) and \(\mathbf{F}_3 = (a\mathbf{i} + b\mathbf{j}) \text{ N}\), where \(a\) and \(b\) are constants. Given that \(P\) is in equilibrium,

1. find the value of \(a\) and the value of \(b\). [2]

The force \(\mathbf{F}_3\) is now removed. The resultant of \(\mathbf{F}_1\) and \(\mathbf{F}_2\) is \(\mathbf{R}\).

1. Find the magnitude of \(\mathbf{R}\). [3]
2. Find the angle, to \(0.1°\), that \(\mathbf{R}\) makes with \(\mathbf{i}\). [2]

1 \includegraphics[max width=\textwidth, alt={}, center]{b9080e9f-2c23-43ce-b171-bd68648dc56b-2_711_398_269_877} Each of two identical light elastic strings has natural length 0.25 m and modulus of elasticity 4 N . A particle \(P\) of mass 0.6 kg is attached to one end of each of the strings. The other ends of the strings are attached to fixed points \(A\) and \(B\) which are 0.8 m apart on a smooth horizontal table. The particle is held at rest on the table, at a point 0.3 m from \(A B\) for which \(A P = B P\) (see diagram).

1. Find the tension in the strings.
2. The particle is released. Find its initial acceleration.

A ship \(S\) is moving with constant velocity \((4\mathbf{i} - 7\mathbf{j})\text{ms}^{-1}\), where \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors due east and due north respectively. Find the speed and direction of \(S\), giving the direction as a three-figure bearing, correct to the nearest degree. [4]

A ship \(S\) is moving with constant velocity \((4\mathbf{i} - 7\mathbf{j})\text{ms}^{-1}\), where \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors due east and due north respectively. Find the speed and direction of \(S\), giving the direction as a three-figure bearing, correct to the nearest degree. [4]

7. [In this question \(\mathbf { i }\) and \(\mathbf { j }\) are horizontal unit vectors due east and due north respectively and position vectors are given relative to a fixed origin \(O\).] Two ships, \(P\) and \(Q\), are moving with constant velocities.
The velocity of \(P\) is \(( 9 \mathbf { i } - 2 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }\) and the velocity of \(Q\) is \(( 4 \mathbf { i } + 8 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }\)

1. Find the direction of motion of \(P\), giving your answer as a bearing to the nearest degree. When \(t = 0\), the position vector of \(P\) is \(( 9 \mathbf { i } + 10 \mathbf { j } ) \mathrm { km }\) and the position vector of \(Q\) is \(( \mathbf { i } + 4 \mathbf { j } ) \mathrm { km }\). At time \(t\) hours, the position vectors of \(P\) and \(Q\) are \(\mathbf { p } \mathrm { km }\) and \(\mathbf { q } \mathrm { km }\) respectively.
2. Find an expression for
   1. \(\mathbf { p }\) in terms of \(t\),
   2. \(\mathbf { q }\) in terms of \(t\).
3. Hence show that, at time \(t\) hours, $$\overrightarrow { Q P } = ( 8 + 5 t ) \mathbf { i } + ( 6 - 10 t ) \mathbf { j }$$
4. Find the values of \(t\) when the ships are 10 km apart.

2. \includegraphics[max width=\textwidth, alt={}, center]{cb92f7b6-2ba5-4703-9595-9ba8570fc52b-04_656_725_283_635} \section*{Figure 1} Figure 1 shows a triangle \(O A C\) where \(O B\) divides \(A C\) in the ratio \(2 : 3\).
Show that \(\mathbf { b } = \frac { 1 } { 5 } ( 3 \mathbf { a } + 2 \mathbf { c } )\)

2. \includegraphics[max width=\textwidth, alt={}, center]{cb92f7b6-2ba5-4703-9595-9ba8570fc52b-04_656_725_283_635} \section*{Figure 1} Figure 1 shows a triangle \(O A C\) where \(O B\) divides \(A C\) in the ratio \(2 : 3\).
Show that \(\mathbf { b } = \frac { 1 } { 5 } ( 3 \mathbf { a } + 2 \mathbf { c } )\)

10. Figure 7 Figure 7 shows a sketch of triangle \(O A B\).
The point \(C\) is such that \(\overrightarrow { O C } = 2 \overrightarrow { O A }\).
The point \(M\) is the midpoint of \(A B\).
The straight line through \(C\) and \(M\) cuts \(O B\) at the point \(N\).
Given \(\overrightarrow { O A } = \mathbf { a }\) and \(\overrightarrow { O B } = \mathbf { b }\)

1. Find \(\overrightarrow { C M }\) in terms of \(\mathbf { a }\) and \(\mathbf { b }\)
2. Show that \(\overrightarrow { O N } = \left( 2 - \frac { 3 } { 2 } \lambda \right) \mathbf { a } + \frac { 1 } { 2 } \lambda \mathbf { b }\), where \(\lambda\) is a scalar constant.
3. Hence prove that \(O N : N B = 2 : 1\)

& \mathbf { r } = \left( \begin{array} { l } a
6
3 \end{array} \right)

3 The vectors \(\mathbf { P } , \mathbf { Q }\) and \(\mathbf { R }\) are given by $$\mathbf { P } = 5 \mathbf { i } + 4 \mathbf { j } , \quad \mathbf { Q } = 3 \mathbf { i } - 5 \mathbf { j } , \quad \mathbf { R } = - 8 \mathbf { i } + \mathbf { j } .$$

1. Find the vector \(\mathbf { P } + \mathbf { Q } + \mathbf { R }\).
2. Interpret your answer to part (i) in the cases
   (A) \(\mathbf { P } , \mathbf { Q }\) and \(\mathbf { R }\) represent three forces acting on a particle,
   (B) \(\mathbf { P } , \mathbf { Q }\) and \(\mathbf { R }\) represent three stages of a hiker's walk.

5. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of a hexagon \(O A B C D E\) where
-the interior angle at \(O\) and at \(C\) are each \(60 ^ { \circ }\) -the interior angle at each of the other vertices is \(150 ^ { \circ }\) -\(O A = O E = B C = C D\) -\(A B = E D = 3 \times O A\) Given that \(\overrightarrow { O A } = \mathbf { a }\) and \(\overrightarrow { O E } = \mathbf { e }\) determine as simplified expressions in terms of \(\mathbf { a }\) and \(\mathbf { e }\)

1. \(\overrightarrow { A B }\)
2. \(\overrightarrow { O D }\) The point \(R\) divides \(A B\) internally in the ratio \(1 : 2\)
3. Determine \(\overrightarrow { R C }\) as a simplified expression in terms of \(\mathbf { a }\) and \(\mathbf { e }\) The line through the points \(R\) and \(C\) meets the line through the points \(O\) and \(D\) at the point \(X\) .
4. Determine \(\overrightarrow { O X }\) in the form \(\lambda \mathbf { a } + \mu \mathbf { e }\) ,where \(\lambda\) and \(\mu\) are real values in simplest form.

A cyclist, when travelling due west at 15 km h\(^{-1}\), finds that the wind appears to be blowing from a bearing of 150°. When the cyclist is travelling due west at 10 km h\(^{-1}\), the wind appears to be blowing from a bearing of 135°. Find the velocity of the wind. [10]

[In this question \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal unit vectors due east and due north respectively.] A man cycling at a constant speed \(u\) on horizontal ground finds that, when his velocity is \(u\mathbf{j}\) m s\(^{-1}\), the velocity of the wind appears to be \(v(3\mathbf{i} - 4\mathbf{j})\) m s\(^{-1}\), where \(v\) is a constant. When the velocity of the man is \(\frac{u}{5}(-3\mathbf{i} + 4\mathbf{j})\) m s\(^{-1}\), he finds that the velocity of the wind appears to be \(w\mathbf{i}\) m s\(^{-1}\), where \(w\) is a constant.

1. Show that \(v = \frac{u}{20}\), and find \(w\) in terms of \(u\). [5]
2. Find, in terms of \(u\), the true velocity of the wind. [2]

A quad-bike, a truck and a car are moving on a large, open, horizontal surface in a desert plain. Relative to the quad-bike, which is travelling due west at its maximum speed of \(10 \text{ m s}^{-1}\), the truck is moving on a bearing of \(340°\). Relative to the car, which is travelling due east at a speed of \(15 \text{ m s}^{-1}\), the truck is moving on a bearing of \(300°\).

1. Show that the speed of the truck is approximately \(24.7 \text{ m s}^{-1}\) and that it is moving on a bearing of \(318°\), correct to the nearest degree. [8 marks]
2. At the instant when the truck is at a distance of \(400\) metres from the quad-bike, the bearing of the truck from the quad-bike is \(060°\). The truck continues to move with the same velocity as in part (a). The quad-bike continues to move at a speed of \(10 \text{ m s}^{-1}\). Find the bearing, to the nearest degree, on which the quad-bike should travel in order to approach the truck as closely as possible. [5 marks]

At noon, a boat \(P\) is on a bearing of \(120°\) from boat \(Q\). Boat \(P\) is moving due east at a constant speed of \(12\) km h\(^{-1}\). Boat \(Q\) is moving in a straight line with a constant speed of \(15\) km h\(^{-1}\) on a course to intercept \(P\). Find the direction of motion of \(Q\), giving your answer as a bearing. [5]

At noon, a boat \(P\) is on a bearing of \(120°\) from boat \(Q\). Boat \(P\) is moving due east at a constant speed of \(12\) km h\(^{-1}\). Boat \(Q\) is moving in a straight line with a constant speed of \(15\) km h\(^{-1}\) on a course to intercept \(P\). Find the direction of motion of \(Q\), giving your answer as a bearing. [5]

4. A rescue boat, whose maximum speed is \(20 \mathrm {~km} \mathrm {~h} ^ { - 1 }\), receives a signal which indicates that a yacht is in distress near a fixed point \(P\). The rescue boat is 15 km south-west of \(P\). There is a constant current of \(5 \mathrm {~km} \mathrm {~h} ^ { - 1 }\) flowing uniformly from west to east. The rescue boat sets the course needed to get to \(P\) as quickly as possible. Find

1. the course the rescue boat sets,
2. the time, to the nearest minute, to get to \(P\). When the rescue boat arrives at \(P\), the yacht is just visible 4 km due north of \(P\) and is drifting with the current. Find
3. the course that the rescue boat should set to get to the yacht as quickly as possible,
4. the time taken by the rescue boat to reach the yacht from \(P\).

1. A \(\operatorname { ship } A\) is moving at a constant speed of \(8 \mathrm {~km} \mathrm {~h} \mathrm {~h} ^ { - 1 }\) on a bearing of \(150 ^ { \circ }\). At noon a second ship \(B\) is 6 km from \(A\), on a bearing of \(210 ^ { \circ }\). Ship \(B\) is moving due east at a constant speed. At a later time, \(B\) is \(2 \sqrt { 3 } \mathrm {~km}\) due south of \(A\).

Find

1. the time at which \(B\) will be due east of \(A\),
2. the distance between the ships at that time.

7. [In this question, the unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are due east and due north respectively. Position vectors are relative to a fixed origin \(O\).] A boat \(P\) is moving with constant velocity \(( - 4 \mathbf { i } + 8 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }\).

1. Calculate the speed of \(P\). When \(t = 0\), the boat \(P\) has position vector \(( 2 \mathbf { i } - 8 \mathbf { j } ) \mathrm { km }\). At time \(t\) hours, the position vector of \(P\) is \(\mathbf { p ~ k m }\).
2. Write down \(\mathbf { p }\) in terms of \(t\). A second boat \(Q\) is also moving with constant velocity. At time \(t\) hours, the position vector of \(Q\) is \(\mathbf { q } \mathrm { km }\), where $$\mathbf { q } = 18 \mathbf { i } + 12 \mathbf { j } - t ( 6 \mathbf { i } + 8 \mathbf { j } )$$ Find
3. the value of \(t\) when \(P\) is due west of \(Q\),
4. the distance between \(P\) and \(Q\) when \(P\) is due west of \(Q\).

1. A \(\operatorname { ship } A\) is moving at a constant speed of \(8 \mathrm {~km} \mathrm {~h} \mathrm {~h} ^ { - 1 }\) on a bearing of \(150 ^ { \circ }\). At noon a second ship \(B\) is 6 km from \(A\), on a bearing of \(210 ^ { \circ }\). Ship \(B\) is moving due east at a constant speed. At a later time, \(B\) is \(2 \sqrt { 3 } \mathrm {~km}\) due south of \(A\).

Find

1. the time at which \(B\) will be due east of \(A\),
2. the distance between the ships at that time.

19

1. (ii) Verify that \(k = 0.8\) [0pt] [1 mark] \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) 19
2. Find the position vector of Amba when \(t = 4\) [0pt] [3 marks] \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) 19
3. At both \(t = 0\) and \(t = 4\) there is a distance of 5 metres between Jo and Amba's positions. Determine the shortest distance between their two parallel lines of motion.
   Fully justify your answer. \includegraphics[max width=\textwidth, alt={}, center]{b7df05bf-f3fc-4705-a13c-6b562896fa9f-32_2492_1721_217_150}

[In this question, \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal unit vectors directed due east and due north respectively and position vectors are given relative to a fixed origin \(O\).] Two ships, \(A\) and \(B\), are moving with constant velocities. The velocity of \(A\) is \((3\mathbf{i} + 12\mathbf{j})\text{ kmh}^{-1}\) and the velocity of \(B\) is \((p\mathbf{i} + q\mathbf{j})\text{ kmh}^{-1}\)

1. Find the speed of \(A\). [2] The ships are modelled as particles. At 12 noon, \(A\) is at the point with position vector \((-9\mathbf{i} + 6\mathbf{j})\) km and \(B\) is at the point with position vector \((16\mathbf{i} + 6\mathbf{j})\) km. At time \(t\) hours after 12 noon, $$\overrightarrow{AB} = [(25 - 12t)\mathbf{i} - 9t\mathbf{j}] \text{ km}$$
2. Find the value of \(p\) and the value of \(q\). [7]
3. Find the bearing of \(A\) from \(B\) when the ships are 15 km apart, giving your answer to the nearest degree. [7]

1. \hspace{0pt} [In this question the horizontal unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are due east and due north respectively.]

A ship \(A\) has constant velocity \(( 4 \mathbf { i } + 2 \mathbf { j } ) \mathrm { kmh } ^ { - 1 }\) and a ship \(B\) has constant velocity \(( - \mathbf { i } + 3 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }\). At noon, the position vectors of the ships \(A\) and \(B\) with respect to a fixed origin \(O\) are \(( - 2 \mathbf { i } + \mathbf { j } ) \mathrm { km }\) and \(( 5 \mathbf { i } - 2 \mathbf { j } ) \mathrm { km }\) respectively. Find

1. the time at which the two ships are closest together,
2. the length of time for which ship \(A\) is within 2 km of ship \(B\).

6. At time \(t\) seconds, where \(t \geqslant 0\), a particle \(P\) moves so that its acceleration \(\mathbf { a } \mathrm { m } \mathrm { s } ^ { - 2 }\) is given by $$\mathbf { a } = 5 t \mathbf { i } - 15 t ^ { \frac { 1 } { 2 } } \mathbf { j }$$ When \(t = 0\), the velocity of \(P\) is \(20 \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }\) Find the speed of \(P\) when \(t = 4\)

1. At time \(t\) seconds, where \(t \geqslant 0\), a particle \(P\) has velocity \(\mathbf { v } \mathrm { ms } ^ { - 1 }\) where

$$\mathbf { v } = \left( t ^ { 2 } - 3 t + 7 \right) \mathbf { i } + \left( 2 t ^ { 2 } - 3 \right) \mathbf { j }$$ Find

1. the speed of \(P\) at time \(t = 0\)
2. the value of \(t\) when \(P\) is moving parallel to \(( \mathbf { i } + \mathbf { j } )\)
3. the acceleration of \(P\) at time \(t\) seconds
4. the value of \(t\) when the direction of the acceleration of \(P\) is perpendicular to \(\mathbf { i }\)

1. In this question you must show all stages of your working.

\section*{Solutions relying entirely on calculator technology are not acceptable.} [In this question, \(\mathbf { i }\) is a unit vector due east and \(\mathbf { j }\) is a unit vector due north.
Position vectors are given relative to a fixed origin \(O\).] At time \(t\) seconds, \(t \geqslant 1\), the position vector of a particle \(P\) is \(\mathbf { r }\) metres, where $$\mathbf { r } = c t ^ { \frac { 1 } { 2 } } \mathbf { i } - \frac { 3 } { 8 } t ^ { 2 } \mathbf { j }$$ and \(c\) is a constant.
When \(t = 4\), the bearing of \(P\) from \(O\) is \(135 ^ { \circ }\)

1. Show that \(c = 3\)
2. Find the speed of \(P\) when \(t = 4\) When \(t = T , P\) is accelerating in the direction of ( \(\mathbf { - i } - \mathbf { 2 7 j }\) ).
3. Find the value of \(T\).

A bee flies in a straight line from \(A\) to \(B\), where \(\overrightarrow{AB} = (3\mathbf{i} - 12\mathbf{j})\) m, in 5 seconds at a constant speed. Find

1. the straight-line distance \(AB\), [2 marks]
2. the speed of the bee, [2 marks]
3. the velocity vector of the bee. [2 marks]

6. At time \(t\) seconds, where \(t \geqslant 0\), a particle \(P\) moves in the \(x - y\) plane in such a way that its velocity \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\) is given by $$\mathbf { v } = t ^ { - \frac { 1 } { 2 } } \mathbf { i } - 4 \mathbf { j }$$ When \(t = 1 , P\) is at the point \(A\) and when \(t = 4 , P\) is at the point \(B\).
Find the exact distance \(A B\).

2 Two horizontal forces of magnitudes 12 N and 19 N act at a point. Given that the angle between the two forces is \(60 ^ { \circ }\), calculate

1. the magnitude of the resultant force,
2. the angle between the resultant and the 12 N force.

\includegraphics{figure_1} A girl swims in still water at 1 m s\(^{-1}\). She swims across a river which is 336 m wide and is flowing at 0.6 m s\(^{-1}\). She sets off from a point \(A\) on one bank and lands at a point \(B\), which is directly opposite \(A\), on the other bank as shown in Fig. 1. Find

1. the direction, relative to the earth, in which she swims, [3]
2. the time that she takes to cross the river. [3]

A river of width 40 m flows with uniform and constant speed between straight banks. A swimmer crosses as quickly as possible and takes 30 s to reach the other side. She is carried 25 m downstream. Find

1. the speed of the river, [2]
2. the speed of the swimmer relative to the water. [2]

1. Two forces \(( 4 \mathbf { i } - 2 \mathbf { j } ) \mathrm { N }\) and \(( 2 \mathbf { i } + q \mathbf { j } ) \mathrm { N }\) act on a particle \(P\) of mass 1.5 kg . The resultant of these two forces is parallel to the vector \(( 2 \mathbf { i } + \mathbf { j } )\).
   1. Find the value of \(q\).
   At time \(t = 0 , P\) is moving with velocity \(( - 2 \mathbf { i } + 4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\).
2. Find the speed of \(P\) at time \(t = 2\) seconds.

2. Ship \(A\) is steaming on a bearing of \(060 ^ { \circ }\) at \(30 \mathrm {~km} \mathrm {~h} ^ { - 1 }\) and at 9 a.m. it is 20 km due west of a second ship \(B\). Ship \(B\) steams in a straight line.

1. Find the least speed of \(B\) if it is to intercept \(A\). Given that the speed of \(B\) is \(24 \mathrm {~km} \mathrm {~h} ^ { - 1 }\),
2. find the earliest time at which it can intercept \(A\).

Two forces \(\mathbf{P}\) and \(\mathbf{Q}\) act on a particle. The force \(\mathbf{P}\) has magnitude \(7\) N and acts due north. The resultant of \(\mathbf{P}\) and \(\mathbf{Q}\) is a force of magnitude \(10\) N acting in a direction with bearing \(120°\). Find

1. the magnitude of \(\mathbf{Q}\),
2. the direction of \(\mathbf{Q}\), giving your answer as a bearing.

[9]

2 A particle has a position vector \(\mathbf { r }\), where \(\mathbf { r } = 4 \mathbf { i } - 5 \mathbf { j }\) and \(\mathbf { i }\) and \(\mathbf { j }\) are unit vectors in the directions east and north respectively.

1. Sketch \(\mathbf { r }\) on a diagram showing \(\mathbf { i }\) and \(\mathbf { j }\) and the origin O .
2. Calculate the magnitude of \(\mathbf { r }\) and its direction as a bearing.
3. Write down the vector that has the same direction as \(\mathbf { r }\) and three times its magnitude.

2 Find the unit vector in the direction of \(\left( \begin{array} { c } 2 \\ - 3 \\ \sqrt { 12 } \end{array} \right)\).

6 \includegraphics[max width=\textwidth, alt={}, center]{8c358a10-a3e1-47b5-ae62-30ba6b76c167-3_655_1011_255_566} The diagram shows triangle \(A B C\) where \(A B = 5 \mathrm {~cm} , A C = 4 \mathrm {~cm}\) and \(B C = 3 \mathrm {~cm}\). Three circles with centres at \(A , B\) and \(C\) have radii \(3 \mathrm {~cm} , 2 \mathrm {~cm}\) and 1 cm respectively. The circles touch each other at points \(E , F\) and \(G\), lying on \(A B , A C\) and \(B C\) respectively. Find the area of the shaded region \(E F G\).

2 \includegraphics[max width=\textwidth, alt={}, center]{f7a22c07-44e3-4891-be60-cbab772f45df-2_549_589_934_778} Two forces, each of magnitude 8 N , act at a point in the directions \(O A\) and \(O B\). The angle between the forces is \(\theta ^ { \circ }\) (see diagram). The resultant of the two forces has component 9 N in the direction \(O A\). Find

1. the value of \(\theta\),
2. the magnitude of the resultant of the two forces.