Vectors – Year 11 GCSE maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. Add the top numbers and the bottom numbers: \(\begin{pmatrix}2\\4\end{pmatrix}\).
   2. Subtract: \(\begin{pmatrix}6\\-6\end{pmatrix}\).
   3. Multiply both numbers by 3: \(\begin{pmatrix}9\\9\end{pmatrix}\).
   4. \(2\mathbf a=\begin{pmatrix}8\\-2\end{pmatrix}\) and \(3\mathbf b=\begin{pmatrix}-6\\15\end{pmatrix}\), so \(2\mathbf a+3\mathbf b=\begin{pmatrix}2\\13\end{pmatrix}\).
   5. \(2\mathbf c=\begin{pmatrix}6\\6\end{pmatrix}\), so \(\mathbf a-2\mathbf c=\begin{pmatrix}4\\-1\end{pmatrix}-\begin{pmatrix}6\\6\end{pmatrix}=\begin{pmatrix}-2\\-7\end{pmatrix}\). Multiply \(\mathbf c\) by 2 first, then subtract each component.

2. Answer to Question 2

   1. \(\sqrt{9+16}=\sqrt{25}=\) 5 Square each component, add, then take the square root. This is Pythagoras' theorem on the arrow of the vector.
   2. \(\sqrt{25+144}=\sqrt{169}=\) 13 The components 5 and 12 form a 5-12-13 right-angled triangle, so the magnitude is 13.
   3. \(\sqrt{64+36}=\sqrt{100}=\) 10 Squaring removes the negative sign: \((-8)^2=64\).
   4. \(\sqrt{4+25}=\sqrt{29}=\) 5.39 29 is not a square number, so round \(\sqrt{29}\) to 2 decimal places.
   5. \(\sqrt{16+49}=\sqrt{65}=\) 8.06 Both components are negative, but the magnitude (a length) is always positive.

3. Answer to Question 3

   1. End minus start: \(\overrightarrow{AB}=\begin{pmatrix}5-1\\5-2\end{pmatrix}=\begin{pmatrix}4\\3\end{pmatrix}\).
   2. \(\overrightarrow{BC}=\begin{pmatrix}-2-5\\4-5\end{pmatrix}=\begin{pmatrix}-7\\-1\end{pmatrix}\) and \(\overrightarrow{CA}=\begin{pmatrix}1-(-2)\\2-4\end{pmatrix}=\begin{pmatrix}3\\-2\end{pmatrix}\).
   3. \(\begin{pmatrix}4\\3\end{pmatrix}+\begin{pmatrix}-7\\-1\end{pmatrix}+\begin{pmatrix}3\\-2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\). The journey \(A\to B\to C\to A\) returns to the start, so the total is the zero vector.
   4. \(\sqrt{16+9}=\) 5 units. The length of \(AB\) is the magnitude of \(\overrightarrow{AB}\), using Pythagoras on its components 4 and 3.

4. Answer to Question 4

   1. Parallel vectors are scalar multiples of each other. \(\mathbf q=3\mathbf p\) and \(\mathbf r=-2\mathbf p\), so \(\mathbf p,\mathbf q\) are parallel, \(\mathbf p,\mathbf r\) are parallel and \(\mathbf q,\mathbf r\) are parallel. \(\mathbf s\) is not parallel to any of them.
   2. \(\mathbf q=3\mathbf p\) because \(3\times2=6\) and \(3\times3=9\). \(\mathbf r=-2\mathbf p\) because \(-2\times2=-4\) and \(-2\times3=-6\).
   3. \(5\times2=10\), so the multiplier is 2 and \(k=4\times2=\) 8.

5. Answer to Question 5

   1. \(\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=-\mathbf a+\mathbf b=\) \(\mathbf b-\mathbf a\). To go from \(A\) to \(B\) we go back along \(\mathbf a\) to \(O\), then along \(\mathbf b\).
   2. \(\overrightarrow{OM}=\tfrac12\mathbf a\) and \(\overrightarrow{ON}=\tfrac12\mathbf b\). A midpoint is halfway along, so we take half of each vector.
   3. \(\overrightarrow{MN}=\overrightarrow{MO}+\overrightarrow{ON}=-\tfrac12\mathbf a+\tfrac12\mathbf b=\tfrac12(\mathbf b-\mathbf a)=\tfrac12\overrightarrow{AB}\). So \(MN\) is parallel to \(AB\) and half its length.

6. Answer to Question 6

   1. \(\overrightarrow{AB}=\begin{pmatrix}8\\10\end{pmatrix}-\begin{pmatrix}2\\1\end{pmatrix}=\begin{pmatrix}6\\9\end{pmatrix}\). Subtract the position vector of the start from the position vector of the end.
   2. \(AP=\tfrac13AB\), so \(\overrightarrow{AP}=\tfrac13\begin{pmatrix}6\\9\end{pmatrix}=\begin{pmatrix}2\\3\end{pmatrix}\). Then \(\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP}=\begin{pmatrix}2\\1\end{pmatrix}+\begin{pmatrix}2\\3\end{pmatrix}=\) \(\begin{pmatrix}4\\4\end{pmatrix}\). \(AP:PB=1:2\) means \(P\) is one third of the way from \(A\) to \(B\). Check: \(\overrightarrow{PB}=\begin{pmatrix}4\\6\end{pmatrix}=2\overrightarrow{AP}\).
   3. Midpoint \(=\tfrac12(\mathbf a+\mathbf b)=\tfrac12\begin{pmatrix}10\\11\end{pmatrix}=\) \(\begin{pmatrix}5\\5.5\end{pmatrix}\). The midpoint is the mean of the two position vectors; add them and halve each component.

7. Answer to Question 7

   1. \(\overrightarrow{WZ}=\begin{pmatrix}3\\2\end{pmatrix}+\begin{pmatrix}-1\\4\end{pmatrix}+\begin{pmatrix}5\\-3\end{pmatrix}=\) \(\begin{pmatrix}7\\3\end{pmatrix}\). The total displacement is the sum of the three movement vectors.
   2. \(\sqrt{7^2+3^2}=\sqrt{58}=\) 7.6 km. The distance is the magnitude of \(\overrightarrow{WZ}\), using Pythagoras.
   3. The return vector is the negative: \(\begin{pmatrix}-7\\-3\end{pmatrix}\). Going back reverses the direction, so both components change sign.
   4. Time \(=\dfrac{7.62}{30}=0.254\) hours \(=0.254\times60=\) 15.2 minutes (about 15 minutes). Time \(=\) distance \(\div\) speed, then multiply by 60 to change hours into minutes.

8. Answer to Question 8

   1. \(\overrightarrow{CH}=\begin{pmatrix}4\\3\end{pmatrix}+\begin{pmatrix}2\\-6\end{pmatrix}=\) \(\begin{pmatrix}6\\-3\end{pmatrix}\). The total displacement is the sum of the two legs.
   2. \(CL=\sqrt{16+9}=5\) km. \(LH=\sqrt{4+36}=\sqrt{40}=6.32\) km. Total \(=5+6.32=\) 11.32 km. Each distance is the magnitude of its vector, found with Pythagoras.
   3. \(CH=\sqrt{36+9}=\sqrt{45}=6.71\) km. The saving is \(11.32-6.71=\) 4.61 km. The direct route is the magnitude of \(\overrightarrow{CH}\); subtract it from the total distance walked.

9. Answer to Question 9

   1. \(\overrightarrow{KT}=\overrightarrow{KO}+\overrightarrow{OT}=\begin{pmatrix}4\\-8\end{pmatrix}+\begin{pmatrix}12\\5\end{pmatrix}=\) \(\begin{pmatrix}16\\-3\end{pmatrix}\). Go from \(K\) back to \(O\) (the negative of \(\overrightarrow{OK}\)), then from \(O\) to \(T\).
   2. \(\sqrt{16^2+3^2}=\sqrt{265}=16.28\), so about 16 paces. The distance is the magnitude of \(\overrightarrow{KT}\).
   3. Midpoint \(=\tfrac12(\begin{pmatrix}12\\5\end{pmatrix}+\begin{pmatrix}-4\\8\end{pmatrix})=\tfrac12\begin{pmatrix}8\\13\end{pmatrix}=\) \(\begin{pmatrix}4\\6.5\end{pmatrix}\). The position vector of a midpoint is the mean of the two position vectors.
   4. \(\sqrt{12^2+5^2}=\sqrt{169}=\) 13 paces. The magnitude of \(\overrightarrow{OT}\) uses the 5-12-13 right-angled triangle.

10. Answer to Question 10

    1. \(\begin{pmatrix}0\\6\end{pmatrix}+\begin{pmatrix}3\\0\end{pmatrix}=\) \(\begin{pmatrix}3\\6\end{pmatrix}\).
    2. \(\sqrt{9+36}=\sqrt{45}=\) 6.7 km/h.
    3. After 1 hour the ferry has gone 6 km north and has drifted 3 km east of the point opposite \(S\).
    4. She needs a sideways component of \(-3\) to cancel the current \(\begin{pmatrix}3\\0\end{pmatrix}\), for example \(\begin{pmatrix}-3\\6\end{pmatrix}\) (and a northward part such as 6).

11. Answer to Question 11

    1. \(3\mathbf x+2\mathbf y=\begin{pmatrix}6\\3\end{pmatrix}+\begin{pmatrix}-2\\6\end{pmatrix}=\) \(\begin{pmatrix}4\\9\end{pmatrix}\).
    2. \(4\mathbf x-\mathbf y=\begin{pmatrix}8\\4\end{pmatrix}-\begin{pmatrix}-1\\3\end{pmatrix}=\) \(\begin{pmatrix}9\\1\end{pmatrix}\).
    3. Across: \(2m-n=7\). Up: \(m+3n=7\). From the first, \(n=2m-7\). Substitute: \(m+3(2m-7)=7\), so \(7m-21=7\), \(7m=28\) and \(m=4\). Then \(n=2\times4-7=\) 1. Check: \(4\mathbf x+\mathbf y=\begin{pmatrix}8\\4\end{pmatrix}+\begin{pmatrix}-1\\3\end{pmatrix}=\begin{pmatrix}7\\7\end{pmatrix}\).

12. Answer to Question 12

    1. \(\begin{pmatrix}120\\50\end{pmatrix}+\begin{pmatrix}-30\\10\end{pmatrix}=\) \(\begin{pmatrix}90\\60\end{pmatrix}\) km/h. Add the plane's own velocity and the wind velocity.
    2. \(3\times\begin{pmatrix}90\\60\end{pmatrix}=\) \(\begin{pmatrix}270\\180\end{pmatrix}\). Multiply the resultant velocity by the time.
    3. \(\sqrt{270^2+180^2}=\sqrt{105300}=\) 324 km to the nearest km. Use Pythagoras on the components of the displacement.
    4. Yes: the position \(\begin{pmatrix}270\\180\end{pmatrix}\) is exactly the destination vector \(\begin{pmatrix}270\\180\end{pmatrix}\), so the wind effect was already allowed for and the plane is on course.

13. Answer to Question 13

    1. \(\overrightarrow{MS}=\begin{pmatrix}34-10\\52-20\end{pmatrix}=\begin{pmatrix}24\\32\end{pmatrix}\) and \(\overrightarrow{SW}=\begin{pmatrix}50-34\\20-52\end{pmatrix}=\begin{pmatrix}16\\-32\end{pmatrix}\).
    2. \(MS=\sqrt{24^2+32^2}=\sqrt{1600}=\) 40.0 m. \(SW=\sqrt{16^2+32^2}=\sqrt{1280}=\) 35.8 m.
    3. \(\dfrac{40}{16}=\) 2.5 seconds. Time \(=\) distance \(\div\) speed.
    4. \(MW=\begin{pmatrix}40\\0\end{pmatrix}\), length \(40\) m. Via \(S\) the ball travels \(40+35.8=75.8\) m. So the direct pass is much shorter.

14. Answer to Question 14

    1. \(\overrightarrow{AB}=\begin{pmatrix}900\\400\end{pmatrix}-\begin{pmatrix}300\\400\end{pmatrix}=\) \(\begin{pmatrix}600\\0\end{pmatrix}\) and \(\overrightarrow{BC}=\begin{pmatrix}900\\1000\end{pmatrix}-\begin{pmatrix}900\\400\end{pmatrix}=\) \(\begin{pmatrix}0\\600\end{pmatrix}\).
    2. \(\overrightarrow{AB}\) is horizontal and \(\overrightarrow{BC}\) is vertical, so the angle at \(B\) is \(90^\circ\) (check: \(600\times0+0\times600=0\)).
    3. \(SA=\sqrt{300^2+400^2}=500\) m, \(AB=600\) m, \(BC=600\) m. Total \(=500+600+600=\) 1 700 m.
    4. \(AC=\sqrt{600^2+600^2}=\sqrt{720000}=849\) m. Via \(B\) it is \(1200\) m, so the saving is \(1200-849=\) 351 m.

15. Answer to Question 15

    1. In a parallelogram \(\overrightarrow{AB}=\overrightarrow{OC}\), so \(\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}=\) \(\mathbf a+\mathbf c\).
    2. \(\overrightarrow{AB}=\mathbf c\), so \(\overrightarrow{OM}=\mathbf a+\tfrac12\mathbf c\). \(\overrightarrow{CB}=\mathbf a\), so \(\overrightarrow{ON}=\mathbf c+\tfrac12\mathbf a\).
    3. \(\overrightarrow{MN}=\overrightarrow{ON}-\overrightarrow{OM}=\mathbf c+\tfrac12\mathbf a-\mathbf a-\tfrac12\mathbf c=\tfrac12\mathbf c-\tfrac12\mathbf a\). Also \(\overrightarrow{AC}=\mathbf c-\mathbf a\). So \(\overrightarrow{MN}=\tfrac12\overrightarrow{AC}\): it is a multiple of \(\overrightarrow{AC}\), so \(MN\) is parallel to \(AC\).
    4. One half.

16. Answer to Question 16

    1. \(\overrightarrow{AB}=\mathbf b-\mathbf a\).
    2. \(AP=\tfrac25AB\), so \(\overrightarrow{AP}=\tfrac25(\mathbf b-\mathbf a)\). Then \(\overrightarrow{OP}=\mathbf a+\tfrac25\mathbf b-\tfrac25\mathbf a=\tfrac35\mathbf a+\tfrac25\mathbf b\). As required.
    3. \(\overrightarrow{OQ}=\tfrac25\mathbf b\). \(\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}=\tfrac25\mathbf b-\tfrac35\mathbf a-\tfrac25\mathbf b=-\tfrac35\mathbf a\).
    4. \(\overrightarrow{PQ}=-\tfrac35\overrightarrow{OA}\), a multiple of \(\overrightarrow{OA}\), so \(PQ\) is parallel to \(OA\) and \(\tfrac35\) of its length (in the opposite direction).

17. Answer to Question 17

    1. \(\overrightarrow{PQ}=(5\mathbf a+4\mathbf b)-(2\mathbf a+\mathbf b)=3\mathbf a+3\mathbf b\). \(\overrightarrow{PR}=(-\mathbf a-2\mathbf b)-(2\mathbf a+\mathbf b)=-3\mathbf a-3\mathbf b\).
    2. \(\overrightarrow{PR}=-1\times\overrightarrow{PQ}\), so the two vectors are parallel. They both start at \(P\), so \(P\), \(Q\) and \(R\) are on the same straight line.
    3. \(PR\) is the same length as \(PQ\) but in the opposite direction, so \(P\) is the midpoint of \(QR\).

18. Answer to Question 18

    1. \(\overrightarrow{BL}=\begin{pmatrix}-6\\8\end{pmatrix}-\begin{pmatrix}12\\16\end{pmatrix}=\) \(\begin{pmatrix}-18\\-8\end{pmatrix}\). Subtract the position vector of \(B\) from that of \(L\).
    2. \(M=\tfrac12(\begin{pmatrix}12\\16\end{pmatrix}+\begin{pmatrix}-6\\8\end{pmatrix})=\tfrac12\begin{pmatrix}6\\24\end{pmatrix}=\) \(\begin{pmatrix}3\\12\end{pmatrix}\). The midpoint is the mean of the two position vectors.
    3. \(CM=\sqrt{9+144}=\sqrt{153}=\) 12.4 km. The distance is the magnitude of \(\overrightarrow{CM}\).
    4. \(\dfrac{12.37}{150}\times60=\) 4.9 minutes. Time \(=\) distance \(\div\) speed, then multiply by 60.
    5. \(V=2\begin{pmatrix}3\\12\end{pmatrix}=\) \((6,24)\). \(M\) is the midpoint of \(CV\). Doubling a vector keeps its direction and doubles its length.

19. Answer to Question 19

    1. \(\overrightarrow{CB}=2\mathbf a\). \(\overrightarrow{OB}=\mathbf c+2\mathbf a\). \(\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\mathbf c+\mathbf a\).
    2. \(\overrightarrow{OD}=\mathbf a+\tfrac12(\mathbf c+\mathbf a)=\tfrac32\mathbf a+\tfrac12\mathbf c\).
    3. \(\overrightarrow{OE}=\tfrac13(2\mathbf a+\mathbf c)=\tfrac23\mathbf a+\tfrac13\mathbf c\). \(\overrightarrow{ED}=\tfrac32\mathbf a+\tfrac12\mathbf c-\tfrac23\mathbf a-\tfrac13\mathbf c=\tfrac56\mathbf a+\tfrac16\mathbf c\) and \(\overrightarrow{EC}=\mathbf c-\tfrac23\mathbf a-\tfrac13\mathbf c=-\tfrac23\mathbf a+\tfrac23\mathbf c\). These are parallel only if \(\tfrac56\div(-\tfrac23)=\tfrac16\div\tfrac23\), which is false (\(-\tfrac54\ne\tfrac14\)). So \(D,E,C\) are not collinear.

20. Answer to Question 20

    1. \(\overrightarrow{OM}=\tfrac12(\mathbf a+\mathbf b)\). \(\overrightarrow{OG}=\tfrac23\overrightarrow{OM}=\tfrac13(\mathbf a+\mathbf b)\).
    2. \(\overrightarrow{ON}=\tfrac12\mathbf a\). \(\overrightarrow{BG}=\tfrac13\mathbf a+\tfrac13\mathbf b-\mathbf b=\tfrac13\mathbf a-\tfrac23\mathbf b\). \(\overrightarrow{BN}=\tfrac12\mathbf a-\mathbf b\). Then \(\overrightarrow{BG}=\tfrac23\overrightarrow{BN}\) because \(\tfrac23\times\tfrac12=\tfrac13\) and \(\tfrac23\times(-1)=-\tfrac23\). So \(BG\) is a multiple of \(BN\) and they share \(B\), so \(B,G,N\) are collinear. \(BG:GN=2:1\).
    3. The median from \(B\) also passes through \(G\) at two thirds of its length. By symmetry the median from \(A\) does too, so all three medians meet at \(G\), the centroid.