Vector Algebra Question 149
Question: 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]
Options:
A) 1 : 1
B) 2 : 5
C) 5 : 2
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- 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} $ .
BETA
