In triangle ABC,if D and E are the mid-points | vedclass

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(A) Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of vertices $A, B, C$ respectively.Since $D$ is the mid-point of $BC$,the position vector

Vedclass · Accepted Answer

$3 \vec{AB}$

Vedclass · Answer

(A) Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of vertices $A, B, C$ respectively.Since $D$ is the mid-point of $BC$,the position vector of $D$ is $\vec{d} = \frac{\vec{b}+\vec{c}}{2}$.Since $E$ is the mid-point of $CA$,the position vector of $E$ is $\vec{e} = \frac{\vec{c}+\vec{a}}{2}$.Now,$\vec{AD} = \vec{d} - \vec{a} = \frac{\vec{b}+\vec{c}}{2} - \vec{a} = \frac{\vec{b}+\vec{c}-2\vec{a}}{2}$.And $\vec{EB} = \vec{b} - \vec{e} = \vec{b} - \frac{\vec{c}+\vec{a}}{2} = \frac{2\vec{b}-\vec{c}-\vec{a}}{2}$.Adding these,$\vec{AD} + \vec{EB} = \frac{\vec{b}+\vec{c}-2\vec{a} + 2\vec{b}-\vec{c}-\vec{a}}{2} = \frac{3\vec{b}-3\vec{a}}{2} = \frac{3}{2}(\vec{b}-\vec{a}) = \frac{3}{2} \vec{AB}$.Therefore,$2(\vec{AD}+\vec{EB}) = 2 	imes \frac{3}{2} \vec{AB} = 3 \vec{AB}$.

In $\triangle ABC$,if $D$ and $E$ are the mid-points of the sides $BC$ and $CA$ respectively,then $2(\vec{AD}+\vec{EB})=$

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