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If $\mathrm{O}$ be the origin and $\mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right), \mathrm{B}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$ are two points,
then what is $(\mathrm{OA})(\mathrm{OB}) \cos \angle \mathrm{AOB}$ ?
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then what is $(\mathrm{OA})(\mathrm{OB}) \cos \angle \mathrm{AOB}$ ?
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Verified Answer
The correct answer is:
$\mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{y}_{1} \mathrm{y}_{2}$
Let $\mathrm{O}(0,0), \mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ and $\mathrm{B}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$ be three points
$\mathrm{OA}=\sqrt{\mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}}, \mathrm{OB}=\sqrt{\mathrm{x}_{2}^{2}+\mathrm{y}_{2}^{2}}$
$\mathrm{AB}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}}$
In $\Delta \mathrm{AOB}$
$\cos \angle \mathrm{AOB}=\frac{\mathrm{OA}^{2}+\mathrm{OB}^{2}-\mathrm{AB}^{2}}{2 . \mathrm{OA} \cdot \mathrm{OB}}$
$\Rightarrow \quad \mathrm{OA} \cdot \mathrm{OB} \cos \angle \mathrm{AOB}=\frac{\mathrm{OA}^{2}+\mathrm{OB}^{2}-\mathrm{AB}^{2}}{2}$
$=\frac{\mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}+\mathrm{x}_{2}^{2}+\mathrm{y}_{2}^{2}-\left\{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}\right\}}{2}$
$\Rightarrow \quad \mathrm{OA} \cdot \mathrm{OB} \cdot \cos \angle \mathrm{AOB}$
$=\frac{\mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}+\mathrm{x}_{2}^{2}+\mathrm{y}_{2}^{2}-\left\{\mathrm{x}_{2}^{2}+\mathrm{x}_{1}^{2}-2 \mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{y}_{2}^{2}+\mathrm{y}_{1}^{2}-2 \mathrm{y}_{1} \mathrm{y}_{2}\right\}}{2}$
$=\frac{2\left(\mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{y}_{1} \mathrm{y}_{2}\right)}{2}=\mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{y}_{1} \mathrm{y}_{2}$
$\mathrm{OA}=\sqrt{\mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}}, \mathrm{OB}=\sqrt{\mathrm{x}_{2}^{2}+\mathrm{y}_{2}^{2}}$
$\mathrm{AB}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}}$
In $\Delta \mathrm{AOB}$
$\cos \angle \mathrm{AOB}=\frac{\mathrm{OA}^{2}+\mathrm{OB}^{2}-\mathrm{AB}^{2}}{2 . \mathrm{OA} \cdot \mathrm{OB}}$
$\Rightarrow \quad \mathrm{OA} \cdot \mathrm{OB} \cos \angle \mathrm{AOB}=\frac{\mathrm{OA}^{2}+\mathrm{OB}^{2}-\mathrm{AB}^{2}}{2}$
$=\frac{\mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}+\mathrm{x}_{2}^{2}+\mathrm{y}_{2}^{2}-\left\{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}\right\}}{2}$
$\Rightarrow \quad \mathrm{OA} \cdot \mathrm{OB} \cdot \cos \angle \mathrm{AOB}$
$=\frac{\mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}+\mathrm{x}_{2}^{2}+\mathrm{y}_{2}^{2}-\left\{\mathrm{x}_{2}^{2}+\mathrm{x}_{1}^{2}-2 \mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{y}_{2}^{2}+\mathrm{y}_{1}^{2}-2 \mathrm{y}_{1} \mathrm{y}_{2}\right\}}{2}$
$=\frac{2\left(\mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{y}_{1} \mathrm{y}_{2}\right)}{2}=\mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{y}_{1} \mathrm{y}_{2}$
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