• # question_answer The points D, E, F divide BC, CA and AB of the triangle ABC in the ratio 1 : 4, 3 : 2 and 3 : 7 respectively and the point K divides AB in the ratio $1:3$, then $(\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF})\,\,:\,\,\overrightarrow{CK}$ is equal to [MNR 1987] A) 1 : 1                                           B) 2 : 5 C) 5 : 2                                           D) None of these

• Let $\overrightarrow{AB}=\mathbf{a},$ $\overrightarrow{AC}=\mathbf{b}$
• So,$\overrightarrow{AD}=\frac{4\mathbf{a}+\mathbf{b}}{5},$$\overrightarrow{AE}=\frac{2\mathbf{b}}{5},$ $\overrightarrow{AF}=\frac{3\mathbf{a}}{10},$ and $\overrightarrow{AK}=\frac{\mathbf{a}}{4}$
• $\frac{\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF}}{\overrightarrow{CK}}=\frac{\frac{\mathbf{b}+4\mathbf{a}}{5}+\frac{2\mathbf{b}-5\mathbf{a}}{5}+\frac{3\mathbf{a}-10\mathbf{b}}{10}}{\frac{\mathbf{a}-4\mathbf{b}}{4}}$
• $=\frac{6\mathbf{b}-2\mathbf{a}+3\mathbf{a}-10\mathbf{b}}{10(\mathbf{a}-4\mathbf{b})}\times 4=\frac{2}{5}$.