Coordinate Geometry Question 61
Question: The distance between the points $ (a\cos \alpha ,a\sin \alpha ) $ and $ (a\cos \beta ,a\sin \beta ) $ is
Options:
A) $ a\cos \frac{\alpha -\beta }{2} $
B) $ 2a\cos \frac{\alpha -\beta }{2} $
C) $ a\sin \frac{\alpha -\beta }{2} $
D) $ 2a\sin \frac{\alpha -\beta }{2} $
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Answer:
Correct Answer: D
Solution:
Distance $ =\sqrt{a^{2}{{(\cos \alpha -\cos \beta )}^{2}}+a^{2}{{(\sin \alpha -\sin \beta )}^{2}}} $
$ =a\sqrt{{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\sin }^{2}}\beta -2\cos \alpha \cos \beta -2\sin \alpha \sin \beta } $
$ =a\sqrt{2{ 1-\cos (\alpha -\beta ) }}=2a\sin ( \frac{\alpha -\beta }{2} ) $
Trick: Put $ a=1,\alpha =\frac{\pi }{2},\beta =\frac{\pi }{6}, $ then the points will be (0, 1) and $ ( \frac{\sqrt{3}}{2},\frac{1}{2} ) $ . Obviously, the distance between these two points is 1 which is given by (d). $ { \because 2a\sin \frac{\alpha -\beta }{2}=2\times 1\times \sin \frac{(\pi /2)-(\pi /6)}{2}=2\times \frac{1}{2}=1 } $
BETA
