已知A(3,0),B(0,3),C(cosa,sina) (1)AC*BC向量=-1,求sina-cosa的值 (2)若\5OA+OC\5向量=√13,且a=(0,π),求OB向量与OC向量的夹角
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![已知A(3,0),B(0,3),C(cosa,sina) (1)AC*BC向量=-1,求sina-cosa的值 (2)若\5OA+OC\5向量=√13,且a=(0,π),求OB向量与OC向量的夹角](/uploads/image/z/5258949-69-9.jpg?t=%E5%B7%B2%E7%9F%A5A%283%2C0%29%2CB%280%2C3%29%2CC%28cosa%2Csina%29+%EF%BC%881%EF%BC%89AC%2ABC%E5%90%91%E9%87%8F%3D-1%2C%E6%B1%82sina-cosa%E7%9A%84%E5%80%BC+%EF%BC%882%EF%BC%89%E8%8B%A5%5C5OA%2BOC%5C5%E5%90%91%E9%87%8F%3D%E2%88%9A13%2C%E4%B8%94a%3D%EF%BC%880%2C%CF%80%EF%BC%89%2C%E6%B1%82OB%E5%90%91%E9%87%8F%E4%B8%8EOC%E5%90%91%E9%87%8F%E7%9A%84%E5%A4%B9%E8%A7%92)
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