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Q. Vector which is perpendicular to $a \, \cos \theta \hat i + b \, \sin \, \theta \hat j$ is

J & K CETJ & K CET 2007Motion in a Plane

Solution:

From definition of dot product of vectors, we have
$ x . y = xy \cos \, \theta$
When $35mm \theta = 90^\circ , cos \, 90^\circ = 0$
$\therefore x . y = 0$
Given,$ x = a \cos \, \theta \, \hat i + b \sin \, \theta \, \hat j$
$ y = b \sin \, \theta \hat i - a \cos \, \theta \hat j$
$ x . y = (a \cos \, \theta \hat i + b \sin \, \theta \hat j) (b \sin \, \theta \hat i - a \cos \, \theta \hat j)$
$ x . y = ab \sin \, \theta \cos \, \theta - ab \sin \, \theta \cos \, \theta = 0$
Hence, vectors are perpendicular.
Similarly for options (b) and (c) also x . y = 0