question_answer

In a right angle \[\Delta ABC,\] the hypotenuse \[AB=p,\] then \[AB\cdot AC+BC\cdot BA+CA\cdot CB\] equals to

A) \[2{{p}^{2}}\]

B) \[{{p}^{2}}\]

C) \[\frac{{{p}^{2}}}{2}\]

D) \[2p\]

Correct Answer: B

Solution :

ABC is right angle triangle AB is hypotenuse and \[AB=p\]

\[\therefore \]\[\angle C=90{}^\circ \]

\[\Rightarrow \]\[\overrightarrow{CA}\cdot \overrightarrow{CB}=0\]\[\Rightarrow \]\[AB\cdot AC=\left| AB \right|\left| AC \right|\cos A\]\[\Rightarrow \]\[BC\cdot BA=\left| BC \right|\left| BA \right|\cos B\]\[\Rightarrow \]\[AB\cdot AC+BC\cdot BA+CA\cdot CB\]

\[\Rightarrow \]\[\left| AB \right|\left| AC \right|\cos A+\left| BC \right|\left| BA \right|\cos B+0\]\[\Rightarrow \]\[\left| AB \right|(\left| AC \right|\cos A+\left| BC \right|\cos B)\]\[\Rightarrow \]\[\left| AB \right|\left| AB \right|={{\left| AB \right|}^{2}}={{p}^{2}}\]

\[[\because p=b\cos B+c\,\cos A]\]