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(D) Let the origin be at $A$. Then $\vec{A} = 0$,$\vec{B} = u$,and $\vec{C} = v$. Since $D$ is the midpoint of $BC$,the position vector of $D$ is $\ve

Vedclass · Accepted Answer

$\frac{|v-3u|}{4}$

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(D) Let the origin be at $A$. Then $\vec{A} = 0$,$\vec{B} = u$,and $\vec{C} = v$. Since $D$ is the midpoint of $BC$,the position vector of $D$ is $\vec{D} = \frac{u+v}{2}$. In $	riangle ABD$,let $M$ be the midpoint of $AD$. The median through vertex $B$ is the segment $BM$. The position vector of $M$ is $\vec{M} = \frac{\vec{A} + \vec{D}}{2} = \frac{0 + \frac{u+v}{2}}{2} = \frac{u+v}{4}$. The vector $\overrightarrow{BM} = \vec{M} - \vec{B} = \frac{u+v}{4} - u = \frac{u+v-4u}{4} = \frac{v-3u}{4}$. The length of the median is $|\overrightarrow{BM}| = |\frac{v-3u}{4}| = \frac{|v-3u|}{4}$.

Let $ABC$ be a triangle. Let $u = \overrightarrow{AB}$ and $v = \overrightarrow{AC}$. If $D$ is the midpoint of $BC$,then the length of the median of $\triangle ABD$ through the vertex $B$ is:

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