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unique2
@unique2
July 2022
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помогите решить 32.
найдите угол между векторами а и 3в-2с, если а (10;10), в(-6;4), с(-9;11)
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mruzunov
|
а|=√х²+у² =√100+100=√200=10√2.
3b=(-18;12),
2с=(-18;22),
3b-2с=(0; -10).
соsα=а·(3b-2с) / (|а|·|3b-2с|)=-100 / 100√2=-√2/2.
α=135°.
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Answers & Comments
3b=(-18;12),
2с=(-18;22),
3b-2с=(0; -10).
соsα=а·(3b-2с) / (|а|·|3b-2с|)=-100 / 100√2=-√2/2.
α=135°.