设平面向量a b的夹角为45度,且|a|=√2 |b|=1 ,若a+ λb与 λa-b的夹角是钝角则实数λ的取值范围为?

2025-06-21 08:59:27
推荐回答(2个)
回答1:

cos(a,b)=a*b/|a|*|b|=√2/2
a*b=√2/2*|a|*|b|=1
因为a+ λb与 λa-b的夹角是钝角

cos(a+λb,λa-b)=(a+λb)(λa-b)/|a+λb||λa-b|<0
则(a+λb)(λa-b)<0
λa^2-ab+λ^2ab-λb^2<0
λ|a|^2+(λ^2-1)ab-λ|b|^2<0
(√2)^2λ+(λ^2-1)*1-λ*1^2<0
2λ+λ^2-1-λ<0
λ^2-λ-1<0
(1-√5)/2<λ<(1+√5)/2

回答2:

cos(a,b)=a*b/|a|*|b|=√2/2
a*b=√2/2*|a|*|b|=1
因为a+ λb与 λa-b的夹角是钝角

cos(a+λb,λa-b)=(a+λb)(λa-b)/|a+λb||λa-b|<0
则(a+λb)(λa-b)<0
λa^2-ab+λ^2ab-λb^2<0
λ|a|^2+(λ^2-1)ab-λ|b|^2<0
(√2)^2λ+(λ^2-1)*1-λ*1^2<0
2λ+λ^2-1-λ<0
λ^2+λ-1<0
(-1-√5)/2<λ<(-1+√5)/2