NIMCET 2014 Question Paper with solutions

Let the position vector of B and C are b and c respectively

Equation of AC
i.e. r = b + λ $\left(\frac{\mathit{b}+\mathit{c}}{4}-\mathit{b}\right)$
and r = 0 + µC
⇒ 1 - $\frac{3\mathit{\lambda }}{4}$ = 0 and $\frac{\mathit{\lambda }}{4}$ = µ
⇒ λ = $\frac{4}{3}$ and µ = $\frac{1}{3}$
Therefore, the positive vector of F is
r = $\frac{1}{3}$ x
AF = $\frac{\mathit{x}}{3}$
⇒ AF = $\frac{1}{3}$ AC
⇒ AF : AC = 1 : 3