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 $\overrightarrow{OP}=\left( 1-k \right)\mathbf{a}\text{+}\frac{k}{2}\left( \mathbf{b}+\mathbf{c} \right)$, where $k$ is a constant and $0<k<1$.
(i)
Show that $A$, $M$ and $P$ are collinear.
[2]
(ii)
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 $\mathbf{c}$.
(iii)
Find $\overrightarrow{AP}\cdot \overrightarrow{BC}$ and state the geometrical meaning of this result.
[4]
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