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1993 Paper 2 Q4
D: 1600.0 B: 1470.3

Two non-parallel lines in 3-dimensional space are given by \(\mathbf{r}=\mathbf{p}_{1}+t_{1}\mathbf{m}_{1}\) and \(\mathbf{r}=\mathbf{p}_{2}+t_{2}\mathbf{m}_{2}\) respectively, where \(\mathbf{m}_{1}\) and \(\mathbf{m}_{2}\) are unit vectors. Explain by means of a sketch why the shortest distance between the two lines is \[ \frac{\left|(\mathbf{p}_{1}-\mathbf{p}_{2})\cdot(\mathbf{m}_{1}\times\mathbf{m}_{2})\right|}{\left|(\mathbf{m}_{1}\times\mathbf{m}_{2})\right|}. \]

  1. Find the shortest distance between the lines in the case \[ \mathbf{p}_{1}=(2,1,-1)\qquad\mathbf{p}_{2}=(1,0,-2)\qquad\mathbf{m}_{1}=\tfrac{1}{5}(4,3,0)\qquad\mathbf{m}_{2}=\tfrac{1}{\sqrt{10}}(0,-3,1). \]
  2. Two aircraft, \(A_{1}\) and \(A_{2},\) are flying in the directions given by the unit vectors \(\mathbf{m}_{1}\) and \(\mathbf{m}_{2}\) at constant speeds \(v_{1}\) and \(v_{2}.\) At time \(t=0\) they pass the points \(\mathbf{p}_{1}\) and \(\mathbf{p}_{2}\), respectively. If \(d\) is the shortest distance between the two aircraft during the flight, show that \[ d^{2}=\frac{\left|\mathbf{p}_{1}-\mathbf{p}_{2}\right|^{2}\left|v_{1}\mathbf{m}_{1}-v_{2}\mathbf{m}_{2}\right|^{2}-[(\mathbf{p}_{1}-\mathbf{p}_{2})\cdot(v_{1}\mathbf{m}_{1}-v_{2}\mathbf{m}_{2})]^{2}}{\left|v_{1}\mathbf{m}_{1}-v_{2}\mathbf{m}_{2}\right|^{2}}. \]
  3. Suppose that \(v_{1}\) is fixed. The pilot of \(A_{2}\) has chosen \(v_{2}\) so that \(A_{2}\) comes as close as possible to \(A_{1}.\) How close is that, if \(\mathbf{p}_{1},\mathbf{p}_{2},\mathbf{m}_{1}\) and \(\mathbf{m}_{2}\) are as in (i)?

1992 Paper 2 Q9
D: 1600.0 B: 1515.9

Let \(\mathbf{a},\mathbf{b}\) and \(\mathbf{c}\) be the position vectors of points \(A,B\) and \(C\) in three-dimensional space. Suppose that \(A,B,C\) and the origin \(O\) are not all in the same plane. Describe the locus of the point whose position vector \(\mathbf{r}\) is given by \[ \mathbf{r}=(1-\lambda-\mu)\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c}, \] where \(\lambda\) and \(\mu\) are scalar parameters. By writing this equation in the form \(\mathbf{r}\cdot\mathbf{n}=p\) for a suitable vector \(\mathbf{n}\) and scalar \(p\), show that \[ -(\lambda+\mu)\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})+\lambda\mathbf{b}\cdot(\mathbf{c}\times\mathbf{a})+\mu\mathbf{c}\cdot(\mathbf{a}\times\mathbf{b})=0 \] for all scalars \(\lambda,\mu.\) Deduce that \[ \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})=\mathbf{b}\cdot(\mathbf{c}\times\mathbf{a})=\mathbf{c}\cdot(\mathbf{a}\times\mathbf{b}). \] Say briefly what happens if \(A,B,C\) and \(O\) are all in the same plane.


Solution: \(\mathbf{r}=(1-\lambda-\mu)\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c} = \mathbf{a} + \lambda(\mathbf{b}-\mathbf{a})+\mu(\mathbf{c}-\mathbf{a})\) Therefore it is the plane through \(\mathbf{a}\) with direction vectors \(\mathbf{b}-\mathbf{a}\) and \(\mathbf{c}-\mathbf{a}\), ie it is the plane through \(\mathbf{a},\mathbf{b},\mathbf{c}\). The normal to this plane will be \((\mathbf{b}-\mathbf{a} ) \times (\mathbf{c}-\mathbf{a}) = \mathbf{b}\times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b}\times \mathbf{a}\), so we must have: \begin{align*} && \mathbf{r} \cdot \left (\mathbf{b}\times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b}\times \mathbf{a} \right) &= \mathbf{a} \cdot \left (\mathbf{b}\times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b}\times \mathbf{a} \right) \\ &&&= \mathbf{a} \cdot (\mathbf{b}\times \mathbf{c}) \end{align*} Therefore, \begin{align*} && \mathbf{a} \cdot (\mathbf{b}\times \mathbf{c}) &= \mathbf{r} \cdot \left (\mathbf{b}\times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b}\times \mathbf{a} \right) \\ &&&= \left ( (1-\lambda-\mu)\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c} \right)\cdot \left (\mathbf{b}\times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b}\times \mathbf{a} \right) \\ &&&= (1-\lambda- \mu) \mathbf{a}\cdot (\mathbf{b} \times \mathbf{c})-\lambda \mathbf{b}\cdot(\mathbf{a} \times \mathbf{c})-\mu \mathbf{c}\cdot(\mathbf{b} \times \mathbf{a}) \\ \Rightarrow && 0 &= (-\lambda- \mu) \mathbf{a}\cdot (\mathbf{b} \times \mathbf{c})-\lambda \mathbf{b}\cdot(\mathbf{a} \times \mathbf{c})-\mu \mathbf{c}\cdot(\mathbf{b} \times \mathbf{a}) \\ &&&= -(\lambda+ \mu) \mathbf{a}\cdot (\mathbf{b} \times \mathbf{c})+\lambda \mathbf{b}\cdot(\mathbf{c} \times \mathbf{a})+\mu \mathbf{c}\cdot(\mathbf{a} \times \mathbf{b}) \\ \end{align*} The result follows from setting \(\mu = 0, \lambda = 1\) and \(\mu = 1, \lambda = 0\). If they all lie in the same plane then the plane described is through the origin, and those values are all the same, but equal to \(0\).