Known A =(1,1), and A and A+2 B in the same direction, A• B.

∵

A =(1,1),

A+2

B=λ

A (λ>0),
∴

B=(λ-1
2,λ-1
2),
∵

A•

B=λ-1
2+λ-1
2=λ-1,
∴

A•

B >-1.

If the module of vector a minus vector b is equal to root number three of 41 minus 20 times root number, the module of vector a is equal to 4, and the module of vector b is equal to 5, then the number product of vectors a and b is? A, b have vector sign,

It is equal to 10√3. The module of vector a minus vector b is equal to the root number three under the root number 41 minus 20 times the root number three. The square of the module of vector a + the square of the module of vector b-2* The quantity product of vectors a and b =41-20√3. The quantity product of vectors a and b is solved as 10√3.

The module of vector a is root number 3, and the module of vector b is 2. Find the module of vector a minus vector a

There's nothing to do. Come and take a look.
(|A-b|)2=a2+b2-2ab
Because |a|=3,|b|=2
A2+ b2=13
∴(|A-b|)2=a2+b2-2ab
=13-2Ab
=13-2|A b cosθ(θ is the angle between a and b)
Cosθ∈[-1,1]
2|A b cos [-12,12]
∴(|A-b|)2∈[1,25]
A-b [1,5]

Let A (-1,0,3), B (0,2,2), C (2,2 C (2,-2,-1), D (1,-1,1) to AB and CD vectors Let A (-1,0,3), B (0,2,2), C (2, C (2,-2,-1), D (1,-1,1) to AB and CD vectors

Vector AB=(1,2,1),
Vector CD=(-1,1,2)
Let n=(x, y, z)
Then n·AB=x+2y+z=0
N·CD=-x+y+2z=0
Let z=0, then x+2y=1
-X+y=-2, x=5/3, y=-1/3, z=1
Now the vector n=(5/3,1/3,1), which is the vector perpendicular to the vectors AB and CD
And then there & amp; apos; s the unit vector that & amp; apos; s perpendicular to AB and CD."
We unitize the vector n, i.e. use the modulus length of n÷n
First calculate the modulus length of n =√(x2+y2+z2), and substitute x=5/3, y=-1/3, z=1 to obtain the modulus length of n =(√35)/3
Module length of n÷n =(5/3÷(√35)/3,1/3÷(√35)/3)
=(√35/7,√35/35,3√35/35)

Let the angle between vector a and b be θ, vector a=(2,1), vector a+2 vector b=(4,5), then cosθ equals?

Let the angle between vector a and b be θ, vector a=(2,1); a+2b=(4,5); then cosθ equals?
Let b=(m, n), then a+2b=(2+2m,1+2n)=(4,5), so 2+2m=4, m=1;1+2n=5, n=2;
Then b=(1,2);
So cos θ=(a▪b)/[ a b ]=(2×1+1×2)/[(√5)(√5)]=4/5.

What happens when two vectors that are not in the same plane are added or multiplied? (Solid space)

Because we mainly study the free vector, so you can let one of the vectors remain unchanged, let the other hetero-plane vector first translation to the plane where the invariant vector is located, and then translation to the same starting point, vector addition or multiplication, especially remind you, vector multiplication is divided into point multiplication and cross multiplication, the specific definition can be entered in Baidu search and understanding of the relevant terms.

As we mainly study the free vector, so you can let one of the vectors remain unchanged, let the other hetero-plane vector first translation to the plane where the invariant vector is located, and then translation to the same starting point, vector addition or multiplication, especially remind you, vector multiplication is divided into point multiplication and cross multiplication, the specific definition can be entered in Baidu search and understanding of the relevant terms.

Known A =(1,1), and A and A+2 B in the same direction, A• B.