If, in a right angled triangle ABC, the hypotenuse AB = p, then AB . AC + BC . BA + CA . CB =

2 p^{2}

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\frac{p^{2}}{2}

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p^{2}

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N o n e

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Solution

The correct option is C $p^{2}$
ABC is a right angled triangle with hypotenuse $A B \Rightarrow ∠ A C B = 90^{\circ}$
$\Rightarrow - - \to A C . - - \to C B = 0 \Rightarrow - - \to C A . - - \to C B = 0$
$- - \to A B . - - \to A C + - - \to B C . - - \to B A + - - \to C A . - - \to C B = - - \to A B . - - \to A C + - - \to C B . - - \to A B + 0$
$= - - \to A B . - - \to A C + - - \to A B . - - \to C B = - - \to A B . (- - \to A C + - - \to C B) = - - \to A B . - - \to A B = | - - \to A B |^{2} = P^{2}$

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- - \to A B . - - \to A C + - - \to B C . - - \to B A + - - \to C A . - - \to C B

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- - \to A B \cdot - - \to A C + - - \to B C \cdot - - \to B A + - - \to C A \cdot - - \to C B

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