2025 EJC P1 Q6

2025 EJC P1 Q6

Junior College 2
9 marks

There are three distinct points \(A\), \(B\) and \(P\). The point \(P\) lies on the circle with centre \(O\) and diameter \(AB\). Relative to the origin \(O\), the position vectors of points \(A\) and \(P\) are \(\mathbf{a}\) and \(\mathbf{p}\) respectively.

  1. By expressing \(\overrightarrow{AP}\) and \(\overrightarrow{BP}\) in terms of \(\mathbf{a}\) and \(\mathbf{p}\), show that \(\overrightarrow{AP}\cdot \overrightarrow{BP}=0\).[3]

The point \(C\) with position vector \(\mathbf{c}\) lies on \(AB\) produced such that \(AC:AB=\lambda :1\).

  1. Find \(\mathbf{c}\) in terms of \(\lambda \) and \(\mathbf{a}\).[2]
  2. It is given that \(PC\) is a tangent to the circle and that angle \(AOP={{120}^{{}^\circ }}\). Using a suitable scalar product, find the value of \(\lambda \).[4]

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Answer:(b) \(\mathbf c=(1-2\lambda)\mathbf a\) (c) \(\lambda=\frac32\)

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