A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{3}\]
C) \[\frac{\pi }{2}\]
D) 0
Correct Answer: C
Solution :
Squaring the given relations implies that \[{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}=0\] Now \[amp\,\,{{z}_{1}}-amp\,\,{{z}_{2}}={{\tan }^{-1}}\frac{{{y}_{1}}}{{{x}_{1}}}-{{\tan }^{-1}}\frac{{{y}_{2}}}{{{x}_{2}}}\] \[={{\tan }^{-1}}\frac{\frac{{{y}_{1}}}{{{x}_{1}}}-\frac{{{y}_{2}}}{{{x}_{2}}}}{1+\frac{{{y}_{1}}{{y}_{2}}}{{{x}_{1}}{{x}_{2}}}}={{\tan }^{-1}}\frac{{{y}_{1}}{{x}_{2}}-{{y}_{2}}{{x}_{1}}}{{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}}\]\[={{\tan }^{-1}}\infty =\frac{\pi }{2}\].
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