Question

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_____

Answer 1

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

Answer 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)

(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

Answer 3

(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

Answer 4

跟三