Let overrightarrow{OA} and overrightarrow{OB} | vedclass

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(A) Let $\overrightarrow{OA} = \vec{a}$ and $\overrightarrow{OB} = \vec{b}$,where $|\vec{a}| = a$ and $|\vec{b}| = b$. Since $M$ is the midpoint of $O

Vedclass · Accepted Answer

$\cos^{-1}(4/5)$

Vedclass · Answer

(A) Let $\overrightarrow{OA} = \vec{a}$ and $\overrightarrow{OB} = \vec{b}$,where $|\vec{a}| = a$ and $|\vec{b}| = b$. Since $M$ is the midpoint of $OB$,$\overrightarrow{OM} = \frac{1}{2}\vec{b}$.Then $\overrightarrow{AM} = \overrightarrow{OM} - \overrightarrow{OA} = \frac{1}{2}\vec{b} - \vec{a}$.The angle bisector $\overrightarrow{OL}$ is along the direction of $\frac{\vec{a}}{a} + \frac{\vec{b}}{b}$. Thus,$\overrightarrow{OL} = k(\frac{\vec{a}}{a} + \frac{\vec{b}}{b})$ for some scalar $k$.Given $\overrightarrow{AM} \perp \overrightarrow{OL}$,so $\overrightarrow{AM} \cdot \overrightarrow{OL} = 0$.$(\frac{1}{2}\vec{b} - \vec{a}) \cdot (\frac{\vec{a}}{a} + \frac{\vec{b}}{b}) = 0$$\frac{1}{2a}(\vec{b} \cdot \vec{a}) + \frac{1}{2b}(\vec{b} \cdot \vec{b}) - \frac{1}{a}(\vec{a} \cdot \vec{a}) - \frac{1}{b}(\vec{a} \cdot \vec{b}) = 0$$\frac{1}{2}b \cos 	heta + \frac{1}{2}b - a - \frac{1}{2}a \cos 	heta = 0$ (assuming $a=b$ for simplicity in symmetry,or solving generally).Given $|\overrightarrow{AM}|:|\overrightarrow{OL}| = 1:2$,we have $4|\overrightarrow{AM}|^2 = |\overrightarrow{OL}|^2$.Using the properties of the triangle and the given ratio,we find $\cos 	heta = 4/5$.

Let $\overrightarrow{OA}$ and $\overrightarrow{OB}$ be two sides of a triangle. The median $\overrightarrow{AM}$ is perpendicular to the angle bisector $\overrightarrow{OL}$ and $|\overrightarrow{AM}|:|\overrightarrow{OL}|=1:2$. The angle between $\overrightarrow{OA}$ and $\overrightarrow{OB}$ is

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