Question

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

• A. $ b\,\sin \,\theta \,\hat{i}-a\,\cos \,\theta \,\hat{j} $
• B. $ \frac{1}{a}\,\sin \,\theta \,\hat{i}-a\,\cos \,\theta \,\hat{j} $
• C. $ 5\hat{k} $
• D. All of the above

Answer 1

The Correct Option is D

From definition of dot product of vectors, we have
$ x\,.\,y=xy\,\,\cos \,\,\theta $
When $ \theta ={{90}^{o}},\,\,\,\cos \,\,{{90}^{o}}=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 option and also
$ x.y=0 $