In \(\triangle ABC\), \(D\) and \(E\) are points on \(BC\) and \(AC\) respectively such that \(BD=2DC\) and \(AE=3EC\). Let \(P\) be the point of intersection of \(AD\) and \(BE\). Find ratio \(\dfrac{BP}{PE}\) using the vector method.

Answer: \(\dfrac{BP}{PE}=\dfrac83\)

Let \(\bar a,\bar b,\bar c,\bar d,\bar e,\bar p\) be the position vectors of \(A,B,C,D,E,P\).

\(BD:DC=2:1\), so \(\bar d=\dfrac{2\bar c+\bar b}{3}\Rightarrow3\bar d=2\bar c+\bar b\) ...(1)

\(AE:EC=3:1\), so \(\bar e=\dfrac{3\bar c+\bar a}{4}\Rightarrow4\bar e=3\bar c+\bar a\) ...(2)

Eliminating \(\bar c\): \(3\times(1)\): \(9\bar d=6\bar c+3\bar b\); \(2\times(2)\): \(8\bar e=6\bar c+2\bar a\).

Subtracting, \(9\bar d-8\bar e=3\bar b-2\bar a\Rightarrow9\bar d+2\bar a=3\bar b+8\bar e\).

Dividing by 11: \(\dfrac{9\bar d+2\bar a}{9+2}=\dfrac{3\bar b+8\bar e}{3+8}\).

The LHS is the position vector of a point on \(AD\) (dividing it in ratio \(9:2\)) and the RHS is that of a point on \(BE\). Since \(AD\) and \(BE\) meet only at \(P\), this common point is \(P\): \(\bar p=\dfrac{8\bar e+3\bar b}{8+3}\).

So \(P\) divides \(BE\) internally in the ratio \(8:3\), i.e. \(\dfrac{BP}{PE}=\dfrac83\).