Let vec{P} = P sin theta hat{i} - P cos theta | vedclass

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(B) Given vector $\vec{P} = P \sin 	heta \hat{i} - P \cos 	heta \hat{j}$.Two vectors are perpendicular if their dot product is zero,i.e.,$\vec{P} \c

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$(Q \cos 	heta \hat{i} + Q \sin 	heta \hat{j})$

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(B) Given vector $\vec{P} = P \sin 	heta \hat{i} - P \cos 	heta \hat{j}$.Two vectors are perpendicular if their dot product is zero,i.e.,$\vec{P} \cdot \vec{Q} = 0$.Let us check option $B$: $\vec{Q} = Q \cos 	heta \hat{i} + Q \sin 	heta \hat{j}$.$\vec{P} \cdot \vec{Q} = (P \sin 	heta \hat{i} - P \cos 	heta \hat{j}) \cdot (Q \cos 	heta \hat{i} + Q \sin 	heta \hat{j})$$= (P \sin 	heta)(Q \cos 	heta) + (-P \cos 	heta)(Q \sin 	heta)$$= PQ \sin 	heta \cos 	heta - PQ \sin 	heta \cos 	heta = 0$.Since the dot product is $0$,the vectors are perpendicular.

Column $I$	Column $II$
$(A) (A+B)$	$(p)$ North-east
$(B) (A-B)$	$(q)$ Vertically upwards
$(C) (A \times B)$	$(r)$ Vertically downwards
$(D) (A \times B) \times (A \times B)$	$(s)$ None

Let $\vec{P} = P \sin \theta \hat{i} - P \cos \theta \hat{j}$ be any vector. Another vector $\vec{Q}$ which is perpendicular to $\vec{P}$ is

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