作业帮已知非零向量a、b,满足a⊥b,且a+2b与a-2b的夹角为120°,则｜a｜/｜b｜等于_____
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已知非零向量a、b,满足a⊥b,且a+2b与a-2b的夹角为120°,则｜a｜/｜b｜等于_____
已知非零向量a、b,满足a⊥b,且a+2b与a-2b的夹角为120°,则｜a｜/｜b｜等于_____
已知非零向量a、b,满足a⊥b,且a+2b与a-2b的夹角为120°,则｜a｜/｜b｜等于_____
因为a⊥b,故a•b=0.
因为a+2b与a-2b的夹角为120°,故cos（a+2b,a-2b）=cos120=-1/2
（a+2b）•（a-2b）/（｜a+2b||a-2b|）=-1/2
(a^2-4b^2)/[根号下(a^2+4a•b+4b^2)•根号下(a^2-4a•b+4b^2)]=-1/2
因为a•b=0,有
(a^2-4b^2)/(a^2+4b^2）=-1/2
a^2=4/3b^2
a^2/b^2=4/3
所以｜a｜/｜b｜=2/根号3
=2√3/3
跟三
(a+2b)(a-2b)=|a|^2-4|b|^2
a⊥b
所以|a+2b|=√(a^2+4b^2+4ab)=√(a^2+4b^2)
|a-2b|=√(a^2+4b^2-4ab)=√(a^2+4b^2)
cos120=-1/2=[|a|^2-4|b|^2]/a^2+4b^2
｜a｜/｜b｜=2
a⊥b => a.b =0
|a+2b|^2 = (a+2b).(a+2b) = |a|^2 +4|b|^2 +4a.b
=> |a+2b| = √(|a|^2 +4|b|^2)
|a-2b|^2 = (a-2b).(a-2b) = |a|^2 +4|b|^2 -4a.b
=> |a-2b| = √(|a|^2 +4|b|^2)
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a⊥b => a.b =0
|a+2b|^2 = (a+2b).(a+2b) = |a|^2 +4|b|^2 +4a.b
=> |a+2b| = √(|a|^2 +4|b|^2)
|a-2b|^2 = (a-2b).(a-2b) = |a|^2 +4|b|^2 -4a.b
=> |a-2b| = √(|a|^2 +4|b|^2)
(a+2b).(a-2b) =|a|^2 -4|b|^2 = |a+2b||a-2b| cos120°
=>|a|^2 -4|b|^2= -1/2(|a|^2 +4|b|^2)
3/2|a|^2 = 2|b|^2
|a|^2/|b|^2 = 4/3
|a|/|b| = 2√3/3
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