Vectors in two dimensions – Year 12 A level maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. Add the \(\mathbf i\) components and the \(\mathbf j\) components separately. \((3-1)\mathbf i+(-2+5)\mathbf j=\) \(2\mathbf i+3\mathbf j\).
   2. Multiply each vector by its scalar first, then subtract component by component. \(2\mathbf a=6\mathbf i-4\mathbf j\) and \(3\mathbf b=-3\mathbf i+15\mathbf j\). So \(2\mathbf a-3\mathbf b=9\mathbf i-19\mathbf j\). So \(2\mathbf a-3\mathbf b=9\mathbf i-19\mathbf j\).
   3. The magnitude of \(p\mathbf i+q\mathbf j\) is \(\sqrt{p^2+q^2}\). \(|\mathbf a|=\sqrt{3^2+(-2)^2}=\) \(\sqrt{13}\). So \(|\mathbf a|=\sqrt{13}\approx3.61\).
   4. A unit vector has magnitude 1, so divide the vector by its own magnitude. \(|\mathbf b|=\sqrt{1+25}=\sqrt{26}\). Unit vector \(=\dfrac1{\sqrt{26}}(-\mathbf i+5\mathbf j)\). Check: its magnitude is \(\dfrac{\sqrt{26}}{\sqrt{26}}=1\) ✓.

2. Answer to Question 2

   1. Use Pythagoras for the magnitude; a sketch shows which quadrant the vector is in. \(\sqrt{25+144}=\) 13. The vector is in the second quadrant: \(\tan^{-1}\left(\tfrac{12}5\right)=67.4^\circ\) above the negative axis, so the angle is \(180^\circ-67.4^\circ=\) 112.6\(^\circ\).
   2. This vector points right and down, so it is in the fourth quadrant. \(\sqrt{16+9}=\) 5. \(\tan^{-1}\left(\tfrac34\right)=36.9^\circ\) below the positive \(x\)-axis, so the angle is \(360^\circ-36.9^\circ=\) 323.1\(^\circ\).
   3. A vector of magnitude \(r\) at angle \(\theta\) to the positive \(x\)-axis is \(r\cos\theta\,\mathbf i+r\sin\theta\,\mathbf j\). \(10\cos30^\circ\mathbf i+10\sin30^\circ\mathbf j=\) \(5\sqrt3\,\mathbf i+5\mathbf j\). Check: \(\sqrt{75+25}=10\) ✓.

3. Answer to Question 3

   1. Parallel vectors are scalar multiples of each other. \(\mathbf p=m\mathbf q\) so \(3=-6m\), \(m=-\tfrac12\). Then \(\lambda=-\tfrac12\times8=\) \(-4\). Check: \(3\mathbf i-4\mathbf j=-\tfrac12(-6\mathbf i+8\mathbf j)\) ✓.
   2. Look for a single number that multiplies both components. \(\begin{pmatrix}-6\\9\end{pmatrix}=-\tfrac32\begin{pmatrix}4\\-6\end{pmatrix}\) because \(-\tfrac32\times4=-6\) and \(-\tfrac32\times(-6)=9\). Parallel, with magnitudes in the ratio \(2:3\). The negative scalar shows that the vectors point in opposite directions.
   3. For parallel vectors the ratios of corresponding components are equal. Parallel means \(\dfrac{2}{k+1}=\dfrac k6\), so \(12=k(k+1)\), \(k^2+k-12=0\), \((k+4)(k-3)=0\). \(k=3\) or \(k=-4\). Check \(k=3\): \(\mathbf a=2\mathbf i+3\mathbf j\) and \(\mathbf b=4\mathbf i+6\mathbf j=2\mathbf a\) ✓.

4. Answer to Question 4

   1. \(\overrightarrow{AB}=\mathbf b-\mathbf a\); its magnitude comes from Pythagoras. \(\overrightarrow{AB}=(8-2)\mathbf i+(-1-5)\mathbf j=6\mathbf i-6\mathbf j\). \(|\overrightarrow{AB}|=\sqrt{72}=\) \(6\sqrt2\). So \(|\overrightarrow{AB}|=6\sqrt2\approx8.49\).
   2. The midpoint has position vector \(\tfrac12(\mathbf a+\mathbf b)\). \(\overrightarrow{OM}=\tfrac12(\mathbf a+\mathbf b)=\tfrac12(10\mathbf i+4\mathbf j)=\) \(5\mathbf i+2\mathbf j\). So \(M\) is the point \((5,2)\).
   3. \(AP:PB=1:2\) means \(P\) is one third of the way from \(A\) to \(B\). \(\overrightarrow{AP}=\tfrac13\overrightarrow{AB}=2\mathbf i-2\mathbf j\). \(\overrightarrow{OP}=2\mathbf i+5\mathbf j+2\mathbf i-2\mathbf j=\) \(4\mathbf i+3\mathbf j\). So \(P\) is the point \((4,3)\).

5. Answer to Question 5

   1. Opposite sides of a parallelogram are equal and parallel, so they are represented by the same vector. \(\overrightarrow{OB}=\mathbf a+\mathbf c\) (since \(\overrightarrow{AB}=\overrightarrow{OC}\)). \(\overrightarrow{AC}=\mathbf c-\mathbf a\).
   2. The midpoint of a line segment has position vector equal to the mean of the end points' position vectors. Midpoint of \(OB\): \(\tfrac12(\mathbf a+\mathbf c)\). Midpoint of \(AC\): \(\mathbf a+\tfrac12(\mathbf c-\mathbf a)=\tfrac12(\mathbf a+\mathbf c)\). The two expressions are identical.
   3. Both midpoints are the same point, so the diagonals of a parallelogram bisect each other. This is a general proof, because \(\mathbf a\) and \(\mathbf c\) can be any two non-parallel vectors.

6. Answer to Question 6

   1. Speed is the magnitude of the velocity; the bearing is measured clockwise from north. Speed \(=\sqrt{9+16}=\) 5 m/s. Bearing \(=\tan^{-1}\left(\tfrac34\right)=\) 036.9\(^\circ\) (east component 3, north 4). The boat moves at 5 m/s on a bearing of \(036.9^\circ\).
   2. With constant velocity, position \(=\) velocity \(\times\) time. \(10\times(3\mathbf i+4\mathbf j)=\) \(30\mathbf i+40\mathbf j\) metres. The boat is 30 m east and 40 m north of the origin, 50 m away.
   3. For a bearing \(\theta\), the east component is \(v\sin\theta\) and the north component is \(v\cos\theta\). East: \(12\sin120^\circ=10.39\). North: \(12\cos120^\circ=-6\). Velocity \(10.39\mathbf i-6\mathbf j\) km/h. The negative \(\mathbf j\) component shows the vessel is heading south of east.

7. Answer to Question 7

   1. Each leg is the difference between consecutive position vectors. Leg 1: \(3\mathbf i+4\mathbf j\). Leg 2: \(5\mathbf i-3\mathbf j\). Leg 3: \(0\mathbf i-6\mathbf j\).
   2. Find the magnitude of each leg and add. Lengths: 5, \(\sqrt{34}=5.83\) and 6. Total \(=\) 16.83 km.
   3. The straight-line displacement is the sum of the three legs. \(\overrightarrow{SF}=8\mathbf i-5\mathbf j\), length \(\sqrt{89}=\) 9.43 km.
   4. Draw a sketch: \(F\) is east and south of \(S\), so the bearing is between \(090^\circ\) and \(180^\circ\). \(F\) is 8 km east and 5 km south. The bearing is \(90^\circ+\tan^{-1}\left(\tfrac58\right)=90^\circ+32.0^\circ=\) \(122^\circ\).

8. Answer to Question 8

   1. The velocity over the ground is the boat's velocity plus the velocity of the current. Velocity \(=3\mathbf i+4\mathbf j\), speed 5 m/s. Time \(=\dfrac{120}4=\) 30 s. Downstream \(=3\times30=\) 90 m.
   2. Measure the bearing clockwise from north using the two components. Bearing \(=\tan^{-1}\left(\tfrac34\right)=\) 036.9\(^\circ\).
   3. To land opposite, the resultant velocity must have no east component. The eastward component must cancel the current: \(4\sin\alpha=3\), \(\sin\alpha=0.75\), \(\alpha=\) 48.6\(^\circ\).
   4. Northward speed \(=4\cos48.6^\circ=\sqrt{16-9}=\sqrt7=2.65\) m/s. Time \(=\dfrac{120}{2.646}=\) 45.4 s. Crossing straight across takes longer, because part of the boat's speed is used against the current.

9. Answer to Question 9

   1. Add the plane's velocity relative to the air and the wind velocity. Ground velocity \(=50\mathbf i+240\mathbf j\). Speed \(=\sqrt{2500+57600}=\) 245.2 km/h. Bearing \(=\tan^{-1}\left(\tfrac{50}{240}\right)=\) 011.8\(^\circ\).
   2. To cancel the eastward wind, the plane must point slightly west of north. The pilot must head west of north: \(240\sin\alpha=50\), \(\alpha=12.0^\circ\). Bearing \(\beta=360^\circ-12.0^\circ=\) 348.0\(^\circ\).
   3. Use Pythagoras: the airspeed is the hypotenuse and the wind is one side. Ground speed \(=\sqrt{240^2-50^2}=\) 234.7 km/h.
   4. Time \(=\dfrac{\text{distance}}{\text{speed}}\); convert the decimal part of an hour to minutes. \(\dfrac{600}{234.7}=2.556\) h \(=\) 2 h 33 min.

10. Answer to Question 10

    1. The velocity is the coefficient of \(t\) in each position vector. \(\mathbf v_P=3\mathbf i+2\mathbf j\) and \(\mathbf v_Q=-\mathbf i+4\mathbf j\).
    2. Substitute \(t=2\) into both position vectors. \(t=2\): \(P=(8,5)\) and \(Q=(12,5)\). Distance \(=\) 4 m.
    3. For a collision both coordinates must be equal at the same time. \(x\): \(2+3t=14-t\) gives \(t=3\). \(y\): \(1+2t=-3+4t\) gives \(t=2\). The times differ, so they are never at the same place at the same time.
    4. Set the \(\mathbf i\) components equal. \(t=3\) (at that time \(P=(11,7)\) and \(Q=(11,9)\), 2 m apart).

11. Answer to Question 11

    1. \(\overrightarrow{AB}=\mathbf b-\mathbf a\). \(\overrightarrow{AB}=(6+4)\mathbf i+(-4-6)\mathbf j=10\mathbf i-10\mathbf j\). So \(\overrightarrow{AB}=10\mathbf i-10\mathbf j\).
    2. \(AP:PB=3:2\) means \(P\) is \(\tfrac35\) of the way from \(A\) to \(B\). \(\overrightarrow{AP}=\tfrac35\overrightarrow{AB}=6\mathbf i-6\mathbf j\). \(\overrightarrow{OP}=(-4+6)\mathbf i+(6-6)\mathbf j=\) \(2\mathbf i\), i.e. \(P=(2,0)\).
    3. Find the magnitude of \(\overrightarrow{AB}\), remembering that each unit is 10 m. \(|\overrightarrow{AB}|=\sqrt{200}=14.14\) tens of metres \(=\) 141 m. The wire is about 141 m long.
    4. Multiply the length by the cost per metre. \(141\times48=\) £6 768 (using 141 m). The wire costs about £6 768.

12. Answer to Question 12

    1. Each pass vector is the difference between the position vectors. \(\overrightarrow{AB}=12\mathbf i+9\mathbf j\) and \(\overrightarrow{BC}=-5\mathbf i+12\mathbf j\).
    2. Use Pythagoras for each magnitude. \(AB=\sqrt{144+81}=15\) m and \(BC=\sqrt{25+144}=13\) m.
    3. Time \(=\dfrac{\text{distance}}{\text{speed}}\). Total distance 28 m. Time \(=\dfrac{28}{15}=\) 1.87 s.
    4. Find \(\overrightarrow{AC}\) and its length, then subtract from the total. \(\overrightarrow{AC}=7\mathbf i+21\mathbf j\), length \(\sqrt{49+441}=\sqrt{490}=22.14\) m. Difference \(=28-22.14=\) 5.86 m.

13. Answer to Question 13

    1. Subtract position vectors. \(\overrightarrow{AB}=3\mathbf i+6\mathbf j\) and \(\overrightarrow{BC}=6\mathbf i+12\mathbf j\).
    2. Three points are collinear when two of the connecting vectors are parallel and share a point. \(\overrightarrow{BC}=2\overrightarrow{AB}\), so the vectors are parallel and share the point \(B\), so \(A\), \(B\), \(C\) lie on a straight line.
    3. Since \(\overrightarrow{BC}=2\overrightarrow{AB}\), \(BC\) is twice as long as \(AB\). \(AB:BC=1:2\).
    4. Find \(|\overrightarrow{AC}|\) and convert units of 100 m to metres. \(\overrightarrow{AC}=9\mathbf i+18\mathbf j\), length \(\sqrt{405}=20.12\) units \(=2\,012\) m. Cost \(=2012\times18=\) £36 216.

14. Answer to Question 14

    1. After 2 hours, \(A\) has travelled 48 km and \(B\) 36 km; resolve using \(\sin\) for east and \(\cos\) for north. \(A\): \(48\sin60^\circ\mathbf i+48\cos60^\circ\mathbf j=41.6\mathbf i+24.0\mathbf j\). \(B\): \(36\sin150^\circ\mathbf i+36\cos150^\circ\mathbf j=18.0\mathbf i-31.2\mathbf j\).
    2. The difference between the bearings is \(150^\circ-60^\circ=90^\circ\). The angle between the two courses is \(90^\circ\), so the distance is \(\sqrt{48^2+36^2}=\) 60 km. (Check: \(\overrightarrow{AB}=-23.6\mathbf i-55.2\mathbf j\), length 60.0.) The ships are 60 km apart.
    3. \(\overrightarrow{AB}=\mathbf r_B-\mathbf r_A\) points south and west, so the bearing is between \(180^\circ\) and \(270^\circ\). \(\overrightarrow{AB}\) points south-west: \(180^\circ+\tan^{-1}\left(\tfrac{23.57}{55.18}\right)=180^\circ+23.1^\circ=\) 203.1\(^\circ\).

15. Answer to Question 15

    1. Opposite sides of a parallelogram give \(\overrightarrow{AB}=\mathbf c\) and \(\overrightarrow{CB}=\mathbf a\). \(\overrightarrow{AB}=\mathbf c\), so \(\overrightarrow{AM}=\tfrac13\mathbf c\) and \(\overrightarrow{OM}=\mathbf a+\tfrac13\mathbf c\). \(\overrightarrow{CB}=\mathbf a\), so \(\overrightarrow{CN}=\tfrac13\mathbf a\) and \(\overrightarrow{ON}=\mathbf c+\tfrac13\mathbf a\).
    2. \(\overrightarrow{MN}=\overrightarrow{ON}-\overrightarrow{OM}=\mathbf c+\tfrac13\mathbf a-\mathbf a-\tfrac13\mathbf c=\tfrac23\mathbf c-\tfrac23\mathbf a=\tfrac23(\mathbf c-\mathbf a)=\tfrac23\overrightarrow{AC}\).
    3. \(\overrightarrow{MN}\) is a scalar multiple of \(\overrightarrow{AC}\), so \(MN\) is parallel to \(AC\) and two thirds of its length.
    4. The scale factor is \(\tfrac23\) (since \(BM=\tfrac23BA\) and \(BN=\tfrac23BC\)), which matches \(MN=\tfrac23AC\).

16. Answer to Question 16

    1. Subtract position vectors: \(\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}\). \(\overrightarrow{PQ}=3\mathbf a+3\mathbf b\) and \(\overrightarrow{PR}=6\mathbf a+6\mathbf b\).
    2. \(\overrightarrow{PR}=2\overrightarrow{PQ}\): parallel vectors with the common point \(P\), so \(P,Q,R\) are collinear.
    3. \(\overrightarrow{QR}=3\mathbf a+3\mathbf b=\overrightarrow{PQ}\), so \(PQ:QR=\) 1:1; \(Q\) is the midpoint of \(PR\).
    4. \(\overrightarrow{PS}\) must be a multiple of \(\mathbf a+\mathbf b\), so the coefficients of \(\mathbf a\) and \(\mathbf b\) must be equal. The line has direction \(\mathbf a+\mathbf b\), so \(\overrightarrow{PS}=(9\mathbf a+(k-1)\mathbf b)=9(\mathbf a+\mathbf b)\), giving \(k-1=9\) and \(k=10\). Check: \(\overrightarrow{OS}=11\mathbf a+10\mathbf b\) continues the pattern \(2,5,8,11\) and \(1,4,7,10\) ✓.

17. Answer to Question 17

    1. The velocity is the coefficient of \(t\); speed is its magnitude. Velocity \(3\mathbf i+2\mathbf j\), speed \(\sqrt{13}=\) 3.61 km/h.
    2. \(\overrightarrow{LB}=\mathbf r-\overrightarrow{OL}\). \(\overrightarrow{LB}=(3t-8)\mathbf i+(2t-1)\mathbf j\). \(|\overrightarrow{LB}|^2=(3t-8)^2+(2t-1)^2=9t^2-48t+64+4t^2-4t+1=13t^2-52t+65\). This is the required expression.
    3. Complete the square to find the minimum of the squared distance. \(13t^2-52t+65=13(t-2)^2+13\). Minimum when \(t=\) 2 h, and the distance is \(\sqrt{13}=\) 3.61 km.
    4. The boat is on the \(x\)-axis when its \(\mathbf j\) component is zero. \(-1+2t=0\) gives \(t=\) 0.5 h. The boat crosses the coastline after half an hour, at the point \((3.5,0)\).

18. Answer to Question 18

    1. For a bearing \(\theta\), east \(=d\sin\theta\) and north \(=d\cos\theta\). \(6\sin250^\circ=6\times(-0.9397)=-5.64\) and \(6\cos250^\circ=6\times(-0.3420)=-2.05\) ✓. Totals: east \(2.50+6.93-5.64=3.79\), north \(4.33-4.00-2.05=-1.72\).
    2. Add the east components and the north components. Displacement \(=3.79\mathbf i-1.72\mathbf j\) km.
    3. Use Pythagoras. Distance \(=\sqrt{3.79^2+1.72^2}=\sqrt{17.32}=\) 4.16 km.
    4. The hiker is south-east of the start. Bearing of hiker from start \(=90^\circ+\tan^{-1}\left(\tfrac{1.72}{3.79}\right)=114.4^\circ\). The return bearing is \(114.4^\circ+180^\circ=\) 294.4\(^\circ\).

19. Answer to Question 19

    1. Each midpoint is the mean of the two end points. \(\mathbf p=\tfrac12(\mathbf a+\mathbf b)\), \(\mathbf q=\tfrac12(\mathbf b+\mathbf c)\), \(\mathbf r=\tfrac12(\mathbf c+\mathbf d)\), \(\mathbf s=\tfrac12(\mathbf d+\mathbf a)\).
    2. \(\overrightarrow{PQ}=\mathbf q-\mathbf p=\tfrac12(\mathbf c-\mathbf a)\). \(\overrightarrow{SR}=\mathbf r-\mathbf s=\tfrac12(\mathbf c+\mathbf d-\mathbf d-\mathbf a)=\tfrac12(\mathbf c-\mathbf a)\).
    3. \(\overrightarrow{PQ}=\overrightarrow{SR}\), so \(PQ\) and \(SR\) are equal and parallel, so \(PQRS\) is a parallelogram.
    4. Similarly \(\overrightarrow{QR}=\tfrac12(\mathbf d-\mathbf b)\). \(PQ=\tfrac12AC\) and \(QR=\tfrac12BD\). In a rectangle the diagonals \(AC\) and \(BD\) are equal, so \(PQ=QR\) and \(PQRS\) is a rhombus.

20. Answer to Question 20

    1. Position \(=\) initial position \(+\) velocity \(\times t\). \(\mathbf r_P=200t\mathbf i+150t\mathbf j\) and \(\mathbf r_Q=(300-100t)\mathbf i+140t\mathbf j\).
    2. \(\overrightarrow{PQ}=\mathbf r_Q-\mathbf r_P=(300-300t)\mathbf i+(140t-150t)\mathbf j=(300-300t)\mathbf i-10t\mathbf j\).
    3. \(|PQ|^2=(300-300t)^2+100t^2=90000-180000t+90000t^2+100t^2=90100t^2-180000t+90000\). The minimum of \(at^2+bt+c\) occurs at \(t=-\dfrac b{2a}=\dfrac{180000}{180200}=0.9989\) h. Minimum value \(=90000-\dfrac{180000^2}{4\times90100}=99.9\), so the minimum distance is \(\sqrt{99.9}\approx\) 10 km.
    4. 10 km is less than 15 km, so the separation rule is broken (at about 59.9 minutes).

21. Answer to Question 21

    1. Position \(=\) initial position \(+\) velocity \(\times t\). \((12-3t)\mathbf i+(16-2t)\mathbf j\). This gives the yacht's position \(t\) hours after 12:00.
    2. The lifeboat travels \(10T\) km in a straight line, so the distance to the interception point equals \(10T\): \((12-3T)^2+(16-2T)^2=100T^2\). Expand: \(144-72T+9T^2+256-64T+4T^2=100T^2\), so \(87T^2+136T-400=0\). So \(a=87\), \(b=136\) and \(c=-400\).
    3. Use the quadratic formula and keep the positive root, since time cannot be negative. \(T=\dfrac{-136+\sqrt{136^2+4\times87\times400}}{2\times87}=\dfrac{-136+\sqrt{157696}}{174}=\dfrac{-136+397.1}{174}=1.50\) h. Interception at about 13:30. Yacht position: \((12-4.50,\,16-3.00)=(7.5,13)\) km. Check: the distance from \(H\) is \(\sqrt{7.5^2+13^2}=15.0\) km \(=10\times1.5\) ✓.