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Distances between Special Points in a Triangle

The points D,...

Question

The points $D, E, F$ divide $- - \to B C, - - \to C A$ and $- - \to A B$ of the triangle $A B C$ in the ratio $1 : 4, 3 : 2$ and $3 : 7$ respectively and the point $K$ divides $- - \to A B$ in the ratio $1 : 3$ , then $(- - \to A D + - - \to B E + - - \to C F) : - - \to C K$ is equal to

1 : 1

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2 : 5

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5 : 2

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3 : 5

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Solution

The correct option is B $2 : 5$
Let $\to a, \to b, \to c$ be the position vectors of vertices $\to A, \to B, \to C$ of the $△ A B C$ . Then the position vectors of $\to D, \to E, \to F$ are respectively
$⎛ ⎝ \frac{4 \to b + \to c}{5} ⎞ ⎠, (\frac{3 \to a + 2 \to c}{5}), ⎛ ⎝ \frac{7 \to a + 3 \to b}{10} ⎞ ⎠$
$∴ - - \to A D = ⎛ ⎝ \frac{4 \to b + \to c}{5} ⎞ ⎠ - \to a$
$- - \to B E = (\frac{3 \to a + 2 \to c}{5}) - \to b$ $- - \to C F = ⎛ ⎝ \frac{7 \to a + 3 \to b}{10} ⎞ ⎠ - \to c$
Now
$- - \to A D + - - \to B E + - - \to C F = \frac{8 \to b + 2 \to c - 10 \to a + 6 \to a + 4 \to c - 10 \to b + 7 \to a + 3 \to b - 10 \to c}{10}$
$= \frac{3 \to a + \to b - 4 \to c}{10}$
$∴_{1} = \frac{3 \to a + \to b - 4 \to c}{10}$
$\Rightarrow 10_{1} = 3 \to a + \to b - 4 \to c$
The position vector of the point $K$ is $\frac{\to b + 3 \to a}{4}$
$∴ - - \to C K = \frac{\to b + 3 \to a}{4} - \to c = \frac{\to b + 3 \to a - 4 \to c}{4}$
$∴_{2} = \frac{\to b + 3 \to a - 4 \to c}{4}$
$∴ 4_{2} = \to b + 3 \to a - 4 \to c$
$∴ 10_{1} = 4_{2}$
$\Rightarrow 5_{1} = 2_{2}$

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