If the vector a b 1 and a b 1,0 , find the coordinates of a and b

Let the vector a be (x1,y1) and b be (x2,y2).

According to |a|=|b|=1, there is x1*x1+y1*y1 =1....1)x2*x2+y2*y2 =1...2)

According to a+b = (1,0).

There is x1+x2 = 1....3)

y1+y2 = 0...4)

The results can be obtained by solving the above four equations together.

You don't have to struggle to solve equations. These two vectors are on the unit circle, and the sum vector is the unit vector on the x-axis, and the drawing comes out.

Let the vector a be (x1,y1) and b be (x2,y2).

According to |a|=|b|=1, there is x1*x1+y1*y1 =1....(1)x2*x2+y2*y2 =1.(2)

According to a+b = (1,0).

There is x1+x2 = 1....(3)

y1+y2 = 0.(4)

Solve the above 4 equations together, and you can stare at the results of the file.

Let the coordinates of a be (x,y)2a+b=(2x+1,2y-3), because it is perpendicular to the regret (2a+b) and qiaodan b, so (2a+b)*b=0(2x+1)*1+[-3*(2y-3)]=02x+1-6y+9=02x-6y=-10x-3y=-5, because a=root number 5, so x 2+y 2=5 x=root number, 5-y, so (filial piety disturbance x-3y) 2=25x 2-6xy+9y 2=25-6xy+5+8y 2=25...

If a b let the vector a=(x,y).

Vector b=(-2,3).

then x=-2 and y=3

x^2y^2=4*13

2 or -2 so a = (-4, 6) or (4, feast -6) because vector a perpendicular vector b

So a*b=0, let a=(x,y).

a*b=02x

3y=0, so x=

Known. Vector a|= 2 13, so town town.

x^2y^2=4*13

then y = 4 or -4

The tomb is named A=(6,4) or.

a=(-6，-4)

Let the key scatter vector b be (Cong Shi x,2x), x 2+(2x) 2=10, and x=root number 2

Therefore, the vector b (root number 2, 2 root number 2).

Let the coordinates of a be (x,y).

By known conditions, we are readily available.

x^2+y^2=3 ①

x+2y=0 ②

From the upper two styles, the pants are easy to obtain.

a(6 5 5,-3 Shi Huai 5 5).

Or. a(-6√5/5,3√5/5)

I don't know how to ask again, hopefully.

Let a=(x,y), then x+2y=0, and then there is |a|=3, so x 2 + y 2 = 9, the simultaneous two formulas can be solved.

Let a=(x,y), then there is x*x+y*y=9, and a is perpendicular b, x+2y=0Solve the equation.

Let a=(x,y), then x 2+y 2=9, and by a b, so 1*x+2*y=0

Combining the above two equations, we get a good ride wheel a=(-6 5,3 5) or (6 5,-3 Youxin 5).

Let the coordinates of vector a be (x,y).

The coordinates of vector b are (m,n).

then x+m=-2

y+n=3x-m=1

y-n=6 is connected: x=-1 2, y=9 2, m=-3 2, n=-3 2, so the coordinates of vector a are (-1 2, 9 2).

Let the vectors a=(m,n) b=(p,q).

then m+p=-2

m-p=1n+q=3

n-q=6 solves this system of equations, and we get m=-1 2, p=-3 2, n=9 2, q=-3 2, so the vector a=(-1 2,9 2) b=(-3 2,-3 2).

Let the coordinates of a be (x,y).

By known conditions, we are readily available.

x^2+y^2=3 ①

x+2y=0 ②

From the above two formulas, it is easy to obtain.

a(6√5/5,-3√5/5)

or a(-6 5 5, 3 5 5).

Let the coordinates of a be (x,y) then:

x^2+y^2=9①

x+2y=0②

Combined with the solution:

I won't count it here, there should be two coordinates)