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 \[AB=p\] |
\[\therefore \]\[\angle C=90{}^\circ \] |
\[\Rightarrow \]\[\overrightarrow{CA}\cdot \overrightarrow{CB}=0\]\[\Rightarrow \]\[AB\cdot AC=\left| AB \right|\left| AC \right|\cos A\]\[\Rightarrow \]\[BC\cdot BA=\left| BC \right|\left| BA \right|\cos B\]\[\Rightarrow \]\[AB\cdot AC+BC\cdot BA+CA\cdot CB\] |
\[\Rightarrow \]\[\left| AB \right|\left| AC \right|\cos A+\left| BC \right|\left| BA \right|\cos B+0\]\[\Rightarrow \]\[\left| AB \right|(\left| AC \right|\cos A+\left| BC \right|\cos B)\]\[\Rightarrow \]\[\left| AB \right|\left| AB \right|={{\left| AB \right|}^{2}}={{p}^{2}}\] |
\[[\because p=b\cos B+c\,\cos A]\] |
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