question_answer

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

A)             \[2{{p}^{2}}\]

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

C)             \[{{p}^{2}}\]

D)             None of these

Correct Answer: C

Solution :

           We have \[\overrightarrow{AB}\,.\,\overrightarrow{AC}+\overrightarrow{BC}\,.\,\overrightarrow{BA}+\overrightarrow{CA}\,.\,\overrightarrow{CB}\]            \[(AB)\,(AC)\cos \theta +(BC)\,(BA)\cos ({{90}^{o}}-\theta )+0\]            =\[AB(AC\cos \theta +BC\sin \theta )=AB\left( \frac{{{(AC)}^{2}}}{AB}+\frac{{{(BC)}^{2}}}{AB} \right)\]                 \[=A{{C}^{2}}+B{{C}^{2}}=A{{B}^{2}}={{p}^{2}}\].