If the scalar product of two vectors is equal to the magnitude of their vector product, find the angle between them.

Since the scalar product of two vectors is equal to the magnitude of their vector product,

\[\overrightarrow{A}.\overrightarrow{B}=|\overrightarrow{A}×\overrightarrow{B}|\]

\[AB\cos\theta=AB\sin\theta\]

\[1=\frac{\sin\theta}{\cos\theta}\]

\[\tan\theta=1\]

\[\therefore\theta=45°\]

Hence, the angle between the two vectors is $45°$.

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