已知向量a=(-3,2)b=(2,1)c=(3,-1)(2)若(a-tb)⊥c求实数t的值

(1)
|a+tb|^2=|a|^2+t^2|b|^2+2t(a.b)
=13+5t^2+2t(-6+2)
=5t^2-8t+13
= 5(t- 4/5)^2 + 49/5
min|a+tb| = 7√5/5
(2)
(a-tb)⊥c
(a-tb).c=0
(-3-2t,2-t).(3,-1)=0
3(-3-2t)-(2-t)=0
9+6t+2-t=0
t=-11/5