已知a+b+c=0,|a|=3,|b|=5,|c|=7 (1)求a与b的夹角θ的余弦值; (2)求实数k,使ka+b与a−2b垂直.
问题描述:
已知
a |
b |
c |
0 |
a |
b |
c |
(1)求
a |
b |
(2)求实数k,使k
a |
b |
a |
b |
答
(1)∵a+b+c=0,∴a+b=−c∴a2+2a•b+b2=c2∵|a|=3,|b|=5,|c|=7∴2a•b=152×3×5cosθ=15∴cosθ=12∴θ=π3(2)(ka+b)⊥ (a−2b)∴(ka+b)• (a−2b)=0即ka2+a•b−2ka•b−2b2=0∵|a|=3...