说明理由1.已知丨向量AB丨=2,丨向量CD=4丨 向量AB与向量CD的夹角为60°,求(1)丨向量AB+向量CD丨(2)向量AB+向量CD与向量AB的夹角
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![说明理由1.已知丨向量AB丨=2,丨向量CD=4丨 向量AB与向量CD的夹角为60°,求(1)丨向量AB+向量CD丨(2)向量AB+向量CD与向量AB的夹角](/uploads/image/z/5165219-11-9.jpg?t=%E8%AF%B4%E6%98%8E%E7%90%86%E7%94%B11.%E5%B7%B2%E7%9F%A5%E4%B8%A8%E5%90%91%E9%87%8FAB%E4%B8%A8%3D2%2C%E4%B8%A8%E5%90%91%E9%87%8FCD%3D4%E4%B8%A8+%E5%90%91%E9%87%8FAB%E4%B8%8E%E5%90%91%E9%87%8FCD%E7%9A%84%E5%A4%B9%E8%A7%92%E4%B8%BA60%C2%B0%2C%E6%B1%82%EF%BC%881%EF%BC%89%E4%B8%A8%E5%90%91%E9%87%8FAB%2B%E5%90%91%E9%87%8FCD%E4%B8%A8%EF%BC%882%EF%BC%89%E5%90%91%E9%87%8FAB%2B%E5%90%91%E9%87%8FCD%E4%B8%8E%E5%90%91%E9%87%8FAB%E7%9A%84%E5%A4%B9%E8%A7%92)
说明理由1.已知丨向量AB丨=2,丨向量CD=4丨 向量AB与向量CD的夹角为60°,求(1)丨向量AB+向量CD丨(2)向量AB+向量CD与向量AB的夹角
说明理由
1.已知丨向量AB丨=2,丨向量CD=4丨 向量AB与向量CD的夹角为60°,
求(1)丨向量AB+向量CD丨
(2)向量AB+向量CD与向量AB的夹角
说明理由1.已知丨向量AB丨=2,丨向量CD=4丨 向量AB与向量CD的夹角为60°,求(1)丨向量AB+向量CD丨(2)向量AB+向量CD与向量AB的夹角
(1)丨向量AB+向量CD丨²=AB²+CD²+2丨AB丨×丨CD丨×cos60°
=4+16+2×2×4×0.5
=24
则丨向量AB+向量CD丨=2根号6
(2)(向量AB+向量CD)·向量AB=AB²+AB·CD
=4²+2×4×0.5
=20
cosθ=(向量AB+向量CD)·向量AB/(丨向量AB+向量CD丨×丨向量AB丨
=20/(2根号6×4)
=(5根号6)/12
θ=arccos(5根号6)/12
1)丨向量AB+向量CD丨²=AB²+CD²+2丨AB丨×丨CD丨×cos60°
=4+16+2×2×4×0.5
=24
则丨向量AB+向量CD丨=2根号6
(2)(向量AB+向量CD)·向量AB=AB...
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1)丨向量AB+向量CD丨²=AB²+CD²+2丨AB丨×丨CD丨×cos60°
=4+16+2×2×4×0.5
=24
则丨向量AB+向量CD丨=2根号6
(2)(向量AB+向量CD)·向量AB=AB²+AB·CD
=4²+2×4×0.5
=20
cosθ=(向量AB+向量CD)·向量AB/(丨向量AB+向量CD丨×丨向量AB丨
=20/(2根号6×4)
=(5根号6)/12
θ=arccos(5根号6)/12
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