Vectors in three dimensions – Year 13 A level maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   Method: work component by component. For \(\mathbf v=x\mathbf i+y\mathbf j+z\mathbf k\), the magnitude is \(|\mathbf v|=\sqrt{x^2+y^2+z^2}\), and the vector from \(A\) to \(B\) is \(\overrightarrow{AB}=\mathbf b-\mathbf a\). Mark scheme as shown in brackets.

   1. Add the components. \(2\mathbf i+2\mathbf j+7\mathbf k\) (1).
   2. Multiply each vector by its scalar first, then subtract component by component. \((9+2)\mathbf i+(-6-8)\mathbf j+(15-4)\mathbf k=11\mathbf i-14\mathbf j+11\mathbf k\) (2).
   3. Square the components, add and take the square root. \(\sqrt{9+4+25}=\sqrt{38}=6.16\) (1). So \(|\mathbf a|=\sqrt{38}\approx6.16\).

2. Answer to Question 2 (4 marks)

   Method: work component by component. For \(\mathbf v=x\mathbf i+y\mathbf j+z\mathbf k\), the magnitude is \(|\mathbf v|=\sqrt{x^2+y^2+z^2}\), and the vector from \(A\) to \(B\) is \(\overrightarrow{AB}=\mathbf b-\mathbf a\). Mark scheme as shown in brackets.

   1. Subtract the coordinates of \(A\) from those of \(B\). \(\overrightarrow{AB}=3\mathbf i+4\mathbf j+12\mathbf k\) (1); \(AB=\sqrt{9+16+144}=13\) (1).
   2. \(\overrightarrow{AC}\) has only a \(\mathbf k\) component, so \(AC\) is its size. \(\overrightarrow{AC}=5\mathbf k\) so \(AC=5\); \(BC=\sqrt{9+16+49}=\sqrt{74}\) (1). So \(BC=\sqrt{74}\approx8.60\).
   3. Use the cosine rule with the three side lengths. \(\cos A=\dfrac{169+25-74}{130}=\dfrac{12}{13}\), \(A=22.6^\circ\) (1). So angle \(BAC\approx22.6^\circ\).

3. Answer to Question 3 (4 marks)

   1. \(\overrightarrow{AC}=6\mathbf i+9\mathbf j+12\mathbf k=3(2\mathbf i+3\mathbf j+4\mathbf k)\) (1). \(\overrightarrow{AB}=2\mathbf i+(p-2)\mathbf j+4\mathbf k\) must be parallel with scale factor 1, so \(p-2=3\) and \(p=5\) (1).
   2. \(\overrightarrow{AB}=2\mathbf i+3\mathbf j+4\mathbf k\) and \(\overrightarrow{BC}=4\mathbf i+6\mathbf j+8\mathbf k=2\overrightarrow{AB}\) (1), so \(AB:BC=1:2\) (1).

   Since \(\overrightarrow{AC}=3(2\mathbf i+3\mathbf j+4\mathbf k)\) and \(\overrightarrow{AB}\) has \(\mathbf i\)-component 2, \(B\) is one third of the way from \(A\) to \(C\). Check: \(\overrightarrow{OB}=(1,2,3)+(2,3,4)=(3,5,7)\) ✓. So \(AB\) is half of \(BC\) in length.

4. Answer to Question 4 (4 marks)

   Method: when \(\mathbf r=\mathbf r_0+t\mathbf v\), the particle moves with constant velocity \(\mathbf v\). Its speed is \(|\mathbf v|\). Mark scheme as shown in brackets.

   1. The velocity is the vector of coefficients of \(t\). Velocity \(=3\mathbf i-2\mathbf j+6\mathbf k\) m/s (1). Speed \(=\sqrt{9+4+36}=7\) m/s (1). So the particle moves at a constant 7 m s\(^{-1}\).
   2. Substitute \(t=5\) and find the magnitude of the position vector. \(\mathbf r=16\mathbf i-8\mathbf j+34\mathbf k\) (1). Distance \(=\sqrt{256+64+1156}=\sqrt{1476}=38.4\) m (1). So after 5 seconds the particle is about 38.4 m from the origin.

5. Answer to Question 5 (4 marks)

   1. \(\mathbf F_1+\mathbf F_2=-4\mathbf i-2\mathbf j+6\mathbf k\), so \(\mathbf F_3=4\mathbf i+2\mathbf j-6\mathbf k\) (1). \(|\mathbf F_3|=\sqrt{16+4+36}=\sqrt{56}=7.48\) N (1).
   2. Resultant \(=-4\mathbf i-2\mathbf j+6\mathbf k\) and \(\mathbf a=\dfrac{1}{2}(-4\mathbf i-2\mathbf j+6\mathbf k)=-2\mathbf i-\mathbf j+3\mathbf k\) m/s\(^2\) (1). \(|\mathbf a|=\sqrt{4+1+9}=\sqrt{14}=3.74\) m/s\(^2\) (1).

   In equilibrium the resultant of all three forces is zero, so \(\mathbf F_3=-(\mathbf F_1+\mathbf F_2)\). Once \(\mathbf F_3\) is removed, the resultant is \(-\mathbf F_3\) and \(\mathbf F=m\mathbf a\) gives the acceleration. Check: \(|\mathbf a|=\dfrac{\sqrt{56}}2=\sqrt{14}\) ✓.

Total : 20 marks