QCM2
ABC est un triangle tel que :
AB = 5, BC = 2 et \(\widehat{ABC} = \frac{2\pi}{3}\), alors \(\overrightarrow{BA} . \overrightarrow{BC} = 5\).
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\(\overrightarrow{BA} . \overrightarrow{BC} = ||\overrightarrow{BA}|| \times ||\overrightarrow{BC}|| \times \cos(\overrightarrow{BA}, \overrightarrow{BC})\)
\(\iff \overrightarrow{BA} . \overrightarrow{BC} = AB \times BC \times \cos\widehat{ABC}\)
\(\iff \overrightarrow{BA} . \overrightarrow{BC} = 5 \times 2 \times \cos\frac{2\pi}{3}\)
\(\iff \overrightarrow{BA} . \overrightarrow{BC} = 10 \times (-\frac{1}{2}) = -5\)