2008 ACJC P2 Q5

2008 ACJC P2 Q5

11 marks
Prelims

The diagram above shows a simple shelter designed by an engineer. It consists of three vertical columns, \(OE\), \(AF\) and \(BG\) with their bases on the horizontal ground. The point \(O\) is taken as the origin and the points \(E\), \(F\) and \(G\) have position vectors given respectively by \(\mathbf{e}=6\mathbf{k}\), \(\mathbf{f}=8\mathbf{j}+10\mathbf{k}\), and \(\mathbf{g}=4\mathbf{i}+2\mathbf{j}+4\mathbf{k}\).

  1. Given that the plane \(EFG\) forms the roof of the shelter, show that the cartesian equation of the plane \(EFG\) is \(3x-2y+4z=24\).[2]
  2. Find the angle between the plane \(EFG\) and the horizontal ground.[2]
  3. The engineer has to drill a hole on the roof of the shelter in order to install a lamp. Given that the lamp is vertically above a point on the ground with position vector \(2\mathbf{i}+3\mathbf{j}\), find the vertical distance from the drilled hole to the horizontal ground.[2]
  4. \(M\) is a point that lies on the line segment \(EF\). The engineer decided to construct a vertical partition that intersects plane \(EFG\) along the line segment \(GM\).Given that the partition has a cartesian equation of \(x+2y=8\), find the vector equation for line \(GM\).[2]
  5. Find the shortest distance from \(A\) to the vertical wall partition.[3]

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Answer:(i) \(3x-2y+4z=24\) (ii) \(42.0^\circ\) (iii) \(6\) units (iv) \(\mathbf r=\begin{pmatrix}8\\0\\0\end{pmatrix}+\alpha\begin{pmatrix}-2\\1\\2\end{pmatrix}\) (v) \(\frac8{\sqrt5}\) units

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