Question

(1)已知a=(3,4),b=(5,12),當(dāng)向量a+kb與a-kb垂直時(shí),求實(shí)數(shù)k的值;

(2)已知a=(3,0),b=(k,5),a與b的夾角為,求實(shí)數(shù)k的值;

(3)已知a=(1,-2),b=(-1,λ),當(dāng)a與b的夾角為鈍角時(shí),求λ的范圍.

Answer 1

(1)解析一:依題

可知a+kb=(3,4)+k(5,12)=(3+5k,4+12k),

a-kb=(3,4)-k(5,12)=(3-5k,4-12k),

由(a+kb)⊥(a-kb),得

(3+5k)(3-5k)+(4+12k)(4-12k)=0,

解得k=±.

解析二:由(a+kb)⊥(a-kb),得

(a+kb)·(a-kb)=0a²-k²b²=0,

即(3²+4²)-k²(5²+12²)=0k=±.

(2)解析:∵a·b=|a||b|cosθ,

∴3×k+0×5=·cos.

解得k=-5.

(3)解析:設(shè)a與b的夾角為θ,

則＜θ＜π,

即cosθ＜0,

且a與b不共線,

則

解得λ＞-且λ≠2,

故λ的取值范圍是(-,2)∪(2,+∞).