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2026 NYGH P1 Q5

2026 NYGH P1 Q5

IP 4
7 marks

In the diagram, \(\overrightarrow{OA}=4\mathbf{b}\), \(\overrightarrow{OC}=6\mathbf{a}\) and \(\overrightarrow{CB}=4\mathbf{a}+\mathbf{b}\). The point \(E\) lies on \(AB\) such that \(AE=EB\).

  1. Express, as simply as possible, in terms of \(\mathbf{a}\) and/or \(\mathbf{b}\),
    1. \(\overrightarrow{AB}\),[1]
    2. \(\overrightarrow{OE}\).[2]
  2. \(G\) is a point on \(AC\) such that \(AG:GC=4:3\). Show that \(O\), \(G\) and \(E\) are collinear.[3]
  3. Find the value of \(\dfrac{\text{area of triangle}\hspace{0.5em}AOG}{\text{area of triangle}\hspace{0.5em}AGE}\).[1]

Solution:

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Answer:(a)(i) \(10\mathbf a-3\mathbf b\) (ii) \(5\mathbf a+\dfrac52\mathbf b\) (b) \(\overrightarrow{OG}=\dfrac{24}{35}\overrightarrow{OE}\), so \(O,G,E\) are collinear (c) \(\dfrac{24}{11}\)

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