Given vector a, vector b is a unit vector,(2 vector a + vector b)·(vector a-2 vector b)=-3√3/2, Ask what the angle between vector a and vector b is

Given vector a, vector b is a unit vector,(2 vector a + vector b)·(vector a-2 vector b)=-3√3/2, Ask what the angle between vector a and vector b is

(2A+B)*(A-2B)=-3√3/2
2A2-2B2-3ab=-3√3/2
-3Ab =-3√3/2
Ab =√3/2
1*1*Cos =√3/2
=30°.

(2A+b)*(a-2b)=-3√3/2
2A^2-2b^2-3ab=-3√3/2
-3Ab=-3√3/2
Ab =√3/2
1*1*Cos=√3/2
=30°.

Given two points A (-2,-3), B (7-3), what is the unit vector of vector AB in the same direction

Because AB vector =(7+2,-3+3)=(9,0)
Then get AB vector // x axis
So unit vector in the same direction as AB is (1,0)

What is the unit vector of vector a=(1.1.-1)?

Divide by his length, which is the root number 3

Given vector a (2,1), find the unit vector perpendicular to vector a

A vector perpendicular to vector a (2,1) is (1,-2)
The modulus of vector (1,-2) is √(1+4)=√5
The unit vector perpendicular to vector a (2,1) is (1/√5)(1,-2) or (-1/√5)(1,-2)
I.e. the unit vector perpendicular to vector a (2,1) is (√5/5,-2√5/5) or (-√5/5,2√5/5)

And vector A =(12,5) Parallel unit vector is () A.(12 13,−5 13) B.(−12 13,−5 13) C.(12 13,5 13) Or (−12 13,−5 13) D.(−12 13,5 13) Or (12 13,−5 13) AND vector A =(12,5) Parallel unit vector is () A.(12 13,−5 13) B.(−12 13,−5 13) C.(12 13,5 13) Or (−12 13,−5 13) D.(−12 13,5 13) Or (12 13,−5 13) AND vector A =(12,5) Parallel unit vector is () A.(12 13,−5 13) B.(−12 13,−5 13) C.(12 13,5 13) Or (−12 13,−5 13) D.(−12) 13,5 13) Or (12 13,−5 13)

Let and vector

A =(12,5) parallel unit vectors

B=(x, y),
|

A|=13 so

A =±13

B


B=(12
13,5
13), Or

B =(−12
13,−5
13)
Therefore, C.

Vector a=(2,1), vector b=(3,4), then what is the projection length of vector a in vector b?

Point A (2,1), B (3,4)
Easy Line OB:4x-3y=0.
|OA |=√(4^2+3^2)=5
The distance from point A to straight line OB is d=|4×2-3×1|/|OA|=5/5=1.
The projection length of vector OA in the vector OB direction is √(|OA|^2-d^2)=2√6, which is the result.