The vectors are known to be a1,1,b1,0,c satisfying ac 0, and ac ,b c 0

a|=|c|= root number 2

A and c vector dots multiply = 0

So vectors a and c are perpendicular.

Vector c = (1,-1) or (-1,1).

But the vector b point multiplied by c is greater than 0, so (-1,1) rounded.

So c=(1,-1).

f:(x,y)→(x',y') = xa+yc, substituting a and c.

xa+yc=（x+y，x-y）=(x',y')=（1，2）x+y=1x-y=2

So x=3 2 , y=-1 2

2)x'^2+y'^2=8

x',y')=(x+2y,x)

So. 2x^2+4xy+4y^2=8

x^2+2xy+2y^2=4

3)cx1,y1)

d(x2,y2)

x1,y1-2)=t(x2,y2-2)

x1=tx2,y1-2=t(y2-2)

Establish. cd:y=kx+2

Substitution. x^2+2xy+2y^2=4

Hit. x^2+2x(kx+2)+2(k^2x^2+4kx+4)=4

1+2k+4k^2)x^2+(4+8k)x+4=0

Discriminant formula = (4+8k) 2-16(1+2k+4k2)=32k>0,k>0

x1+x2=-(4+8k)/(1+2k+4k^2)=(1+t)x2,x1*x2=4/(1+2k+4k^2)=t

x2^24/(1+2k+4k^2)=[t/(1+t)^2][16(1+2k)^2/(1+2k+4k^2)^2

1+t)^2/t=4(1+2k)^2/(1+2k+4k^2)=4(1+2k/(1+2k+4k^2)=4+8/(1/k+4k+2)>4

and <=4+8 (2+2sqr(4)=16 3

Solution. t>0,t≠1

1 3 so. 1/3< t<3 and t≠1< p>

Let c=xa+yb, then (-1,-2)=(x,x)+(y,-y)=(x+y,x-y), i.e., {x+y= -1,x-y= -2, the manuscript is honored with x = 3 2, y = 1 2, so c = 3 2*a+1 2*b(mainly using the square mask cavity group) key stuffy section.

a+b+c=0

c=-a-b

c|^2=(-a-b)^2=|a|^2+2ab+|b|2 Because the ascending segment a b is violently carried by ab=0

c|^2=(-a-b)^2=|a|^2+2ab+|b|^2=1+0+4=5

So noisy|c|= root number 5

Let c=(x,y) then the answer is missing c+b=(x+1,y+2) c-a=(x-1,y+1) because (c+b) a, (c-a) smooth track b

Therefore, x+1-y-2=0,2x-2-y-1=0 is solved to x=2,y=1 and the argument is c=(2,1).

Let c=ax+by

Then: x+y=-1

x-y=2 to solve the equation y=-3 2

x=1 2So: Feast Blind Disturbance Dawn C=1 2A-3 2B

Because a b, so.

For a+b+c=0, multiply a by the same time as du, then 丨zhia丨 +, thus 1+, i.e.

(a-b) daoc, so.

And because the return is 0=(a+b+c)(a+b+c)=丨Aa丨 +丨b丨 +丨c丨 +

丨a丨 +丨b丨 +丨c丨 +

So 丨a丨 +丨b丨 +丨c丨 =

How can three squares be a negative number, get 4, and finally the algebra is miscalculated.