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Physics Question on Motion in a plane

Vector which is perpendicular to a cos⁡θi^+b sin⁡ θj^a \, \cos \theta \hat i + b \, \sin \, \theta \hat j is

A

bsin⁡ θi^−a cos⁡ θj^b \sin \, \theta \hat i - a \, \cos \, \theta \hat j

B

1asin⁡ θi^−1bcos⁡ θj^\frac{1}{a} \sin \, \theta \hat i - \frac{1}{b} \cos \, \theta \hat j

C

5k^5 \hat k

D

All of these

Answer

All of these

Explanation

Solution

From definition of dot product of vectors, we have
x.y=xycos⁡ θx . y = xy \cos \, \theta
When 35mmθ=90∘,cos 90∘=035mm \theta = 90^\circ , cos \, 90^\circ = 0
∴x.y=0\therefore x . y = 0
Given,x=acos⁡ θ i^+bsin⁡ θ j^ x = a \cos \, \theta \, \hat i + b \sin \, \theta \, \hat j
y=bsin⁡ θi^−acos⁡ θj^y = b \sin \, \theta \hat i - a \cos \, \theta \hat j
x.y=(acos⁡ θi^+bsin⁡ θj^)(bsin⁡ θi^−acos⁡ θj^)x . y = (a \cos \, \theta \hat i + b \sin \, \theta \hat j) (b \sin \, \theta \hat i - a \cos \, \theta \hat j)
x.y=absin⁡ θcos⁡ θ−absin⁡ θcos⁡ θ=0x . 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