2022 NYJC J2 CT2 P1 Q6

With reference to the origin \(O\), the non-collinear points \(A\),\(B\) and \(C\) are such that \(\overrightarrow{OA}=\mathbf{a}\), \(\overrightarrow{OB}=\mathbf{b}\) and \(\overrightarrow{OC}=\mathbf{c}\). The \(M\) is the midpoint of \(BC\). The point \(P\) is such that \(\overrightarrow{OP}=\left( 1-k \right)\mathbf{a}\mathrm{+}\frac{k}{2}\left( \mathbf{b}+\mathbf{c} \right)\), where \(k\) is a constant and \(0<k<1\).

1. Show that \(A\), \(M\) and \(P\) are collinear.[2]
2. Show that the area of triangle \(BCP\) is \(\frac{\lambda }{2}\left| \mathbf{a}\times \mathbf{b}+\mathbf{b}\times \mathbf{c}-\mathbf{a}\times \mathbf{c} \right|\) where \(\lambda \), a constant in terms of \(k\), is to be found.[2]

It is now given that \(\left| \mathbf{b} \right|=\left| \mathbf{c} \right|\) and the angle between vectors \(\mathbf{a}\) and \(\mathbf{b}\) is equal to the angle between vectors \(\mathbf{a}\) and \(\mathbf{c}\).

3. Find \(\overrightarrow{AP}\cdot \overrightarrow{BC}\) and state the geometrical meaning of this result.[4]