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Vector Addition

In a triangle...

Question

In a triangle $A B C$ , if two velocity vectors are represented in magnitude and direction by the vectors $\to D C$ and $\to B E$ where $D$ and $E$ are the mid points of the sides $A B$ and $A C$ respectively. Then the resultant of the two velocities will be

A

\to B C

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B

\frac{3}{2} \to B C

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C

\frac{1}{2} \to B C

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D

\frac{2}{3} \to B C

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Solution

The correct option is B $\frac{3}{2} \to B C$
Let this has vectors ofthe vertices A,B,C
Let $\to a, \to b, \to c$ respectively
$O A, \to a, O B, \to b, O C, \to c$
Since, D is the point of AB
$- - \to O D = \frac{\to a + \to b}{2}$
Similarly $- - \to O E = \frac{\to a + \to b}{2} N o w, - - \to B E = - - \to O E - - - \to O B = \frac{\to a + \to c}{2} - \to b = \frac{\to a + \to b - 2 \to b}{2}$
$- - \to D C = - - \to O C - - - \to O D = \to c - \frac{\to a + \to b}{2} = \frac{2 \to c - \to a - \to b}{2} - - \to B E + - - \to D C = \frac{\to a + \to c - 2 \to b}{2} + \frac{2 \to c - \to a - \to b}{2} = \frac{\to a + \to c - 2 \to b + 2 \to c - \to a - \to b}{2} = \frac{3 \to c - 3 \to b}{2} = \frac{3}{2} (\to c - \to b) N o w, - - \to B C = - - \to O C - - - \to O B = \to c - \to b - - \to B E + - - \to D C = \frac{3}{2} - - \to B C$

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