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

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(C) Let the triangle be right-angled at $C$. Thus,$\overrightarrow{CA} \cdot \overrightarrow{CB} = 0$ because $\overrightarrow{CA} \perp \overrightarr

Vedclass · Accepted Answer

$p^2$

Vedclass · Answer

(C) Let the triangle be right-angled at $C$. Thus,$\overrightarrow{CA} \cdot \overrightarrow{CB} = 0$ because $\overrightarrow{CA} \perp \overrightarrow{CB}$.We need to evaluate $S = \overrightarrow{AB} \cdot \overrightarrow{AC} + \overrightarrow{BC} \cdot \overrightarrow{BA} + \overrightarrow{CA} \cdot \overrightarrow{CB}$.Since $\overrightarrow{CA} \cdot \overrightarrow{CB} = 0$,the expression simplifies to $S = \overrightarrow{AB} \cdot \overrightarrow{AC} + \overrightarrow{BC} \cdot \overrightarrow{BA}$.Using the dot product definition $\vec{u} \cdot \vec{v} = |u||v| \cos 	heta$:$\overrightarrow{AB} \cdot \overrightarrow{AC} = |AB||AC| \cos 	heta = p \cdot |AC| \cdot \frac{|AC|}{p} = |AC|^2$.$\overrightarrow{BC} \cdot \overrightarrow{BA} = |BC||BA| \cos(90^o - 	heta) = |BC| \cdot p \cdot \frac{|BC|}{p} = |BC|^2$.Therefore,$S = |AC|^2 + |BC|^2$.By Pythagoras theorem,$|AC|^2 + |BC|^2 = |AB|^2 = p^2$.Thus,the value is $p^2$.

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

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