Question

设向量a,b是两个不共线的非零向量.

(1)若a与b起点相同,当t为何值时,a,tb,1/3(a+b)三个向量的终点在一条直线上?
(2)若|a|=|b|,且a与b的夹角为60度,那么当t为何值时,|a-tb|的值最小?
求过程!!

Answer 1

(1)
(1/3)(a+b) = k(a+tb)
=> 1/3 = k and 1/3 = kt
=> t=1
t=1, a,tb,(1/3)(a+b)三个向量的终点在一条直线上
(2)
|a-tb|^2
=|a|^2+t^2|b|^2-2t|a||b|cos60度
=|a|^2(t^2-t+1)
=|a|^2( (t-1/2)^2 +3/4)
min |a-tb| at t=1/2
min |a-tb| = (√3/2)|a|