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

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1. Answer to Question 1 (4 marks)

   1. Multiply each vector by its scalar, then subtract component by component. \(12\mathbf i-9\mathbf j-2\mathbf i-4\mathbf j=10\mathbf i-13\mathbf j\) (2) So \(3\mathbf a-2\mathbf b=10\mathbf i-13\mathbf j\).
   2. Find \(\mathbf a+2\mathbf b\) first, then use Pythagoras for the magnitude. \(\mathbf a+2\mathbf b=6\mathbf i+\mathbf j\) (1), magnitude \(\sqrt{37}=6.08\) (1) So \(|\mathbf a+2\mathbf b|=\sqrt{37}\approx6.08\).

   Check (b) with components: \((4+2)\mathbf i+(-3+4)\mathbf j=6\mathbf i+\mathbf j\) ✓. Writing \(\sqrt{6^2+1}\) as 7 is a frequent arithmetic slip.

2. Answer to Question 2 (4 marks)

   1. Parallel vectors are scalar multiples, so the ratios of corresponding components are equal. \(\dfrac{k+2}2=\dfrac93=3\) (1), \(k=4\) (1) Check: \(\mathbf p=6\mathbf i+9\mathbf j=3(2\mathbf i+3\mathbf j)\) ✓.
   2. If \(\mathbf p=k\mathbf q\) with \(k\gt 0\), then \(|\mathbf p|=k|\mathbf q|\). \(\mathbf p=3\mathbf q\) (1) so the ratio is \(3:1\) (1) So \(\mathbf p\) is three times as long as \(\mathbf q\).

   Equivalently \(3(k+2)=2\times9\), giving \(k+2=6\).

3. Answer to Question 3 (4 marks)

   1. \(\overrightarrow{AB}=\mathbf b-\mathbf a\); its magnitude comes from Pythagoras. \(6\mathbf i-6\mathbf j\) (1), \(6\sqrt2\) (1) So \(|\overrightarrow{AB}|=6\sqrt2\approx8.49\).
   2. Ratio \(2:1\) means \(P\) is \(\tfrac23\) of the way from \(A\) to \(B\). \(\overrightarrow{AP}=\tfrac23(6\mathbf i-6\mathbf j)=4\mathbf i-4\mathbf j\) (1), \(\overrightarrow{OP}=5\mathbf i+0\mathbf j\) (1) So \(P\) is the point \((5,0)\).

   Alternatively \(\overrightarrow{OP}=\tfrac13\mathbf a+\tfrac23\mathbf b=\tfrac13(1,4)+\tfrac23(7,-2)=(5,0)\) ✓. Taking \(\tfrac13\) of \(\overrightarrow{AB}\) gives \((3,2)\), the point dividing \(AB\) in the ratio \(1:2\).

4. Answer to Question 4 (4 marks)

   1. For a bearing \(\theta\), the east component is \(v\sin\theta\) and the north component is \(v\cos\theta\). \(20\sin150^\circ=10\) and \(20\cos150^\circ=-17.32\), so \(10\mathbf i-17.3\mathbf j\) (2) The ship heads south-east, so the \(\mathbf j\) component is negative.
   2. With constant velocity, displacement \(=\) velocity \(\times\) time. \(30\mathbf i-52.0\mathbf j\) (1) After 3 hours the ship is 30 km east and 52 km south of its start.
   3. Read off the \(\mathbf i\) component of the position vector. 30 km (1) The ship is 30 km east of its starting point.

5. Answer to Question 5 (4 marks)

   Write \(\overrightarrow{OM}=\tfrac12\mathbf a\) and \(\overrightarrow{ON}=\tfrac12\mathbf b\), then find \(\overrightarrow{MN}\) and compare it with \(\overrightarrow{AB}\).

   \(\overrightarrow{AB}=\mathbf b-\mathbf a\) (1). \(\overrightarrow{MN}=\overrightarrow{ON}-\overrightarrow{OM}=\tfrac12\mathbf b-\tfrac12\mathbf a\) (1) \(=\tfrac12(\mathbf b-\mathbf a)=\tfrac12\overrightarrow{AB}\) (1). A scalar multiple, so parallel and half the length (1).

   This is the midpoint theorem: the line joining the midpoints of two sides of a triangle is parallel to the third side and half as long.

Total : 20 marks