If D, E, F are mid-points of the sides BC, CA and AB respectively of $∆ A B C, then \vec{A D} + \vec{B E} + \vec{C F} =______________________.$

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Solution

Given:
D, E, F are mid-points of the sides BC, CA and AB respectively

Let $\vec{a}, \vec{b} and \vec{c}$ are the position vectors of A, B and C, respectively.

$Now, \vec{B D} = \frac{1}{2} \vec{B C} = \frac{1}{2} (\vec{c} - \vec{b}) (∵ D is the mid - point of B C) Thus, in ∆ A B D, using triangle law of vector addition, \vec{A D} = \vec{A B} + \vec{B D} \Rightarrow \vec{A D} = \vec{b} - \vec{a} + \frac{1}{2} \vec{c} - \frac{1}{2} \vec{b} \Rightarrow \vec{A D} = \frac{1}{2} \vec{b} + \frac{1}{2} \vec{c} - \vec{a} . . . (1) Similarly, \vec{C E} = \frac{1}{2} \vec{C A} = \frac{1}{2} (\vec{a} - \vec{c}) (∵ E is the mid - point of A C) Thus, in ∆ B C E, using triangle law of vector addition, \vec{B E} = \vec{B C} + \vec{C E} \Rightarrow \vec{B E} = \vec{c} - \vec{b} + \frac{1}{2} \vec{a} - \frac{1}{2} \vec{c} \Rightarrow \vec{B E} = \frac{1}{2} \vec{c} + \frac{1}{2} \vec{a} - \vec{b} . . . (2) And \vec{A F} = \frac{1}{2} \vec{A B} = \frac{1}{2} (\vec{b} - \vec{a}) (∵ F is the mid - point of A B) Thus, in ∆ A C F, using triangle law of vector addition, \vec{C F} = \vec{C A} + \vec{A F} \Rightarrow \vec{C F} = \vec{a} - \vec{c} + \frac{1}{2} \vec{b} - \frac{1}{2} \vec{a} \Rightarrow \vec{C F} = \frac{1}{2} \vec{b} + \frac{1}{2} \vec{a} - \vec{c} . . . (3) Adding (1), (2) and (3), we get \vec{A D} + \vec{B E} + \vec{C F} = \frac{1}{2} \vec{b} + \frac{1}{2} \vec{c} - \vec{a} + \frac{1}{2} \vec{c} + \frac{1}{2} \vec{a} - \vec{b} + \frac{1}{2} \vec{b} + \frac{1}{2} \vec{a} - \vec{c} = \vec{0}$

Hence, $\vec{A D} + \vec{B E} + \vec{C F} = \vec{0} .$

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