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In a right-angled triangle $\mathrm{ABC}$, if the hypotenuse $\mathrm{AB}=\mathrm{p}$, then what is $\overrightarrow{\mathrm{AB}}$. $\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{BC}} . \overrightarrow{\mathrm{BA}}+\overrightarrow{\mathrm{CA}} \cdot \overrightarrow{\mathrm{CB}}$ equal to?
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The correct answer is:
$\mathrm{p}^{2}$
$\begin{aligned} \overrightarrow{A B} \cdot \overrightarrow{A C}+\overrightarrow{B C} \cdot \overrightarrow{B A}+\overrightarrow{C A} \cdot \overrightarrow{C B} \\=&(A B \cdot A C \cdot \cos \theta)+(B C \cdot B A \cdot \cos (90-\theta)) \\ & \quad+(C A \cdot C B \cdot \cos 90) \\=&(p \cdot x \cdot \cos \theta)+(y \cdot p \cdot \sin \theta)+0 \\=& p[x \cos \theta+y \sin \theta] \end{aligned}$
By projection formula :
$p=x \cos \theta+y \cos (90-\theta)$
$=x \cos \theta+y \sin \theta$
$\therefore \quad p[x \cos \theta+y \sin \theta]=p \times p=p^{2} .$
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