向量的數學問題~

Question

設向量a=(1,2).向量b=(1,-1)
1.若t為實數,試求|向量a+t向量b|的最小值.
2.若(x向量a+y向量b)⊥向量a,且|x向量a+y向量b|=3,試求x及y的值.
第一題~給我答案就好了!!
第二題~麻煩給我算式!!
各位高手~幫我一下吧

Answer 1

1. |向量a+t向量b|
=|(1,2)+t(1,-1)|
=|(1+t,2-t)|
=√[(1+t)^2+(2-t)^2]
=√(2t^2-2t+5)
=√[2(t^2-t)+5]
=√[2(t-1/2)^2-1/2+5]
=√[2(t-1/2)^2+9/2]
所以，當t=1/2時，有最小值√(9/2)= 3/√2。

2. (x向量a+y向量b)⊥向量a
=> [x(1,2)+y(1,-1)]•(1,2)=0
=> (x+y, 2x-y)•(1,2)=0
=> x+y+4x-2y=0
=> 5x-y=0 ..........(1)
|x向量a+y向量b|=3
=> |x(1,2)+y(1,-1)|=3
=> |(x+y, 2x-y)|=3
=> (x+y)^2+(2x-y)^2=3^2=9
=> 5x^2-2xy+2y^2=9 ........(2)
由(1)得 y=5x 代入(2)
5x^2-10x^2+50x^2=9
=> 45x^2=9
=> x=±1/√5
y=5x=±√5

Answer 2

你們都好強喔

Answer 3

1.3√2/2
2.(x向量a+y向量b)⊥向量a=>(x+y,2x-y)⊥(1,2)=x+y+4x-2y=5x-y=>y=5x......<1>
|x向量a+y向量b|=3=>(x+y)^2+(2x-y)^2=3^2=>5x^2-2xy+2y^2=9........<2>
把<1>帶入<2>=>x^2=1/5=>x=√5/5
由<1>知道y=5√5/5