Question

已知向量a=（-3,2）,b=（2,1）,c=（3,1）,t∈R

Answer 1

(1)

a-tb=(-3-2t,2-t)

|a-tb|^2

=(-3-2t)^2+(2-t)^2

=5t^2+8t+13

(|a-tb|^2)' = 10t+8 =0

t=-4/5

(|a-tb|^2)''=10>0 (min)

min|a-tb|^2 = 5(-4/5)^2+8(-4/5)+13 = 16/5-32/5+13=49/5

min|a-tb| = 7√5/5

(2)

a+tb // c

-(3+2t)/(2-t)=3/1

-3-2t=6-3t

t=9