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

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1. Answer to Question 1 (4 marks)

   1. \(\begin{pmatrix}4\\6\end{pmatrix}-\begin{pmatrix}-1\\5\end{pmatrix}=\) \(\begin{pmatrix}5\\1\end{pmatrix}\) (1 for \(2\mathbf a\), 1 for the answer) Multiply each component of \(\mathbf a\) by 2, then subtract the components of \(\mathbf b\); subtracting \(-1\) adds 1.
   2. \(\mathbf a+\mathbf b=\begin{pmatrix}1\\8\end{pmatrix}\), magnitude \(\sqrt{1+64}=\sqrt{65}=\) 8.06 (1 for the vector, 1 for the answer) Add the vectors first, then square the components, add them and take the square root.

   Check (a) by components: \(2\times2-(-1)=5\) and \(2\times3-5=1\).

2. Answer to Question 2 (4 marks)

   1. \(\begin{pmatrix}6\\-2\end{pmatrix}+\begin{pmatrix}-10\\5\end{pmatrix}=\) \(\begin{pmatrix}-4\\3\end{pmatrix}\) (2) The total displacement is the sum of the two legs: add the top numbers and the bottom numbers.
   2. \(\sqrt{16+9}=\) 5.0 km (1 for the sum of squares, 1 for the answer) The distance is the magnitude of \(\overrightarrow{PR}\). The components \(-4\) and 3 make a 3-4-5 triangle, so \(PR=5.0\) km.

   A common error in (b) is to subtract the squares, \(\sqrt{16-9}\).

3. Answer to Question 3 (4 marks)

   1. \(\mathbf r=-2\mathbf p\) because \(-2\times3=-6\) and \(-2\times-2=4\) (2) Two vectors are parallel when one is a multiple of the other. Here every component of \(\mathbf r\) is \(-2\) times the matching component of \(\mathbf p\).
   2. \(9=3\times3\), so \(k=3\times(-2)=\) \(-6\) (1 for the multiplier, 1 for \(k\)) For \(\mathbf q\) to be parallel to \(\mathbf p\), it must be a multiple of \(\mathbf p\). The top numbers show the multiplier is 3, so the bottom number must also be multiplied by 3.

4. Answer to Question 4 (4 marks)

   1. \(\overrightarrow{AB}=\mathbf b-\mathbf a\) (1) Go from \(A\) back to \(O\), then along \(\mathbf b\) to \(B\).
   2. \(\overrightarrow{AN}=\tfrac14(\mathbf b-\mathbf a)\) (1), so \(\overrightarrow{ON}=\mathbf a+\tfrac14\mathbf b-\tfrac14\mathbf a=\tfrac34\mathbf a+\tfrac14\mathbf b\) (1) \(AN:NB=1:3\) means \(N\) is a quarter of the way from \(A\) to \(B\).
   3. \(\overrightarrow{MN}=\overrightarrow{ON}-\overrightarrow{OM}=\tfrac34\mathbf a+\tfrac14\mathbf b-\tfrac12\mathbf a=\tfrac14\mathbf a+\tfrac14\mathbf b\) (1) \(\overrightarrow{OM}=\tfrac12\mathbf a\) because \(M\) is the midpoint of \(OA\).

   In (c), \(\overrightarrow{MN}=\tfrac14(\mathbf a+\mathbf b)\), so \(MN\) is parallel to the line joining \(O\) to the midpoint of \(AB\).

5. Answer to Question 5 (4 marks)

   1. \(\begin{pmatrix}24\\10\end{pmatrix}+\begin{pmatrix}-8\\-16\end{pmatrix}=\) \(\begin{pmatrix}16\\-6\end{pmatrix}\) (1) The displacement from \(B\) to \(R\) is the sum of the two legs.
   2. \(BS=\sqrt{576+100}=26\), \(SR=\sqrt{64+256}=17.9\), total 43.9 km (1 for the two lengths, 1 for the total) Find the magnitude of each leg with Pythagoras and add them.
   3. Direct return \(=\sqrt{256+36}=17.1\) km. \(43.9+17.1=61.0\) km, which is more than 60 km, so no (1) The direct return distance is the magnitude of \(\overrightarrow{BR}\).

Total : 20 marks