Given vector a=(-1,2), the vector is a unit vector parallel to vector a, find vector b.

Given vector a=(-1,2), the vector is a unit vector parallel to vector a, find vector b.

Let vector b be (x, y)
Then X2+Y2=1;
X/-1=Y/2
Solution X =5/5
Y=±2√5/5
So vector b (√5/5,-2√5/5) and vector b (-√5,2√5)
Hope it works for you.

Let b be (x, y)
Then X2+Y2=1;
X/-1=Y/2
Solution X =5/5
Y=±2√5/5
So vector b (√5/5,-2√5/5) and vector b (-√5,2√5)
Hope it works for you.

How to Find the Projection of Vector a on Vector b

Multiply the modulus of vector a by the cosine of the angle formed by the two vectors
|A|*cos

For a non-zero vector a b, if a parallel b is the projection of a in the direction of b | a| For a non-zero vector a b, if a parallel b is the condition that a is projected into |a| in the direction of b

A is projected in the direction of b as |a|
So │a│cosθ=│a│
Cos θ=1θ=0
Therefore a parallel b
Therefore, a parallel b is a necessary and insufficient condition for the projection of a in the direction of b to be |a|

A is projected in the direction of b as |a|
So │a│cosθ=│a│
Cos θ=1θ=0
Therefore a parallel b
Therefore, a parallel b is a necessary and insufficient condition for projection of a in the direction of b to be |a|

If vector a =(-1,2),|b|=3 root number 5, and vector a·b=-|a b|, then vector b=() A,(-3,6) B.(3,-6) C,(6,3) D,(-6,3) Note: The above letters are in bold (i.e. vector)

Vector a·b=-|a||b|
The angle between a and b is 180 degrees, i.e. a and b are collinear in the opposite direction.
Set b=ma, m <0
B2= m2a2
I.e.45=5 m2
Get: m2=9
Because m <0
So, m=-3 So, b=-3a=(3,-6)
Option B
Have fun! Hope to help you, if not understand, please ask, wish to learn progress! O (∩_∩) O

Vector a·b=-|a||b|
Then: a, b angle is 180 degrees, i.e. a, b reverse collinear
Set b=ma, m <0
B2= m2a2
I.e.45=5 m2
Get: m2=9
Because m <0
So, m=-3 So, b=-3a=(3,-6)
Option B
Have fun! Hope to help you, if not understand, please ask, wish to learn progress! O (∩_∩) O

Vector a·b=-|a||b|
Then: a, b angle is 180 degrees, i.e. a, b reverse collinear
Set b=ma, m <0
B2= m2a2
I.e.45=5 m2
Get: m2=9
Because m <0
So, m=-3 So, b=-3a=(3,-6)
Select B
Have fun! Hope to help you, if not understand, please ask, wish to learn progress! O (∩_∩) O

Let a=(2,0), b (root number 3,1), the result is? Let a=(2,0), b (root 3,1), the result is?

1. Modulus equality, i.e.|a|=|b|;
2. The included angle is 30°.

1. Modularity, i.e.|a|=|b|;
2. The included angle is 30°.

Given vector a=(-root number 3,1) vector b=(1,-root number 3), find < vector a, vector b >

Solution
/A /=√(-√3)2+1 2=2
/B /=√1 2+(-√3)=2
Ab =(-√3)×1+1×(-√3)=-2√3
∴
Cos=ab/|a||b|
=-2√3/(2×2)
=-√3/2
∵∈[0,π]
∴=150