If in a right-angled triangle ABC,the hypoten | vedclass

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(A) Let the vertices of the triangle be $A, B, C$. Given that the triangle is right-angled at $C$,we have $\overrightarrow{CA} \perp \overrightarrow{C

Vedclass · Accepted Answer

$p^2$

Vedclass · Answer

(A) Let the vertices of the triangle be $A, B, C$. Given that the triangle is right-angled at $C$,we have $\overrightarrow{CA} \perp \overrightarrow{CB}$,which implies $\overrightarrow{CA} \cdot \overrightarrow{CB} = 0$. We know that $\overrightarrow{AB} = \overrightarrow{AC} + \overrightarrow{CB}$,so $\overrightarrow{AC} = \overrightarrow{AB} - \overrightarrow{CB}$. Substituting this into the expression: $E = \overrightarrow{AB} \cdot \overrightarrow{AC} + \overrightarrow{BC} \cdot \overrightarrow{BA} + \overrightarrow{CA} \cdot \overrightarrow{CB}$ $E = \overrightarrow{AB} \cdot (\overrightarrow{AB} - \overrightarrow{CB}) + (-\overrightarrow{CB}) \cdot (-\overrightarrow{AB}) + 0$ $E = |\overrightarrow{AB}|^2 - \overrightarrow{AB} \cdot \overrightarrow{CB} + \overrightarrow{CB} \cdot \overrightarrow{AB}$ Since $\overrightarrow{AB} \cdot \overrightarrow{CB} = \overrightarrow{CB} \cdot \overrightarrow{AB}$,these terms cancel out. $E = |\overrightarrow{AB}|^2 = p^2$.

If in a right-angled triangle $ABC$,the hypotenuse $|\overrightarrow{AB}| = p$,then $\overrightarrow{AB} \cdot \overrightarrow{AC} + \overrightarrow{BC} \cdot \overrightarrow{BA} + \overrightarrow{CA} \cdot \overrightarrow{CB} = $

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