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已知向量a=(-3,2),b=(2,1),c=(3,1),t∈R
1个回答
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(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
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
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