If overrightarrow{P} cdot overrightarrow{Q} = | vedclass

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(A) The dot product of two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ is defined as $\overrightarrow{P} \cdot \overrightarrow{Q} = PQ \cos

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$0$

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(A) The dot product of two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ is defined as $\overrightarrow{P} \cdot \overrightarrow{Q} = PQ \cos 	heta$,where $	heta$ is the angle between the two vectors.Given that $\overrightarrow{P} \cdot \overrightarrow{Q} = PQ$.Substituting the definition,we get $PQ \cos 	heta = PQ$.Dividing both sides by $PQ$ (assuming $P, Q 
eq 0$),we get $\cos 	heta = 1$.Since $\cos 0^\circ = 1$,the angle $	heta$ must be $0^\circ$.

If $\overrightarrow{P} \cdot \overrightarrow{Q} = PQ,$ then the angle between $\overrightarrow{P}$ and $\overrightarrow{Q}$ is ....... $^\circ$.

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