How is the sign representation of the unit vector represented, such as the unit vector of vector a, is it plus a ^ ah? How is the sign representation of the unit vector represented, such as the unit vector of vector a is not added with a ^ ah?

How is the sign representation of the unit vector represented, such as the unit vector of vector a, is it plus a ^ ah? How is the sign representation of the unit vector represented, such as the unit vector of vector a is not added with a ^ ah?

The unit vector doesn't have a special symbol. So, there's no such thing as a representation. And there's no such thing as a unit vector for vector a.
If in a spatial rectangular coordinate system,
The unit vector which is the same as the positive direction of x-axis, y-axis and z-axis is generally represented by vector i, vector j and vector k.

The unit vector doesn't have a special sign. So, there's no such thing as a representation. And there's no such thing as a unit vector for vector a.
If in a spatial rectangular coordinate system,
The unit vector which is the same as the positive direction of x-axis, y-axis and z-axis is generally represented by vector i, vector j and vector k.

The vectors a, b are all unit vectors, and the vector a. is =1/2, the angle between vector a-c and b-c is π/6, then the absolute value of vector a-c The vectors a, b are all unit vectors, and the vector a.b=1/2, the angle between vector a-c and b-c is π/6, then the maximum absolute value of vector a-c is ()

A. b=1/2=|a b|* cos (X)= cos (X)=> X=60 so that the unit vectors a, b have an angle of 60 degrees
Let a be OA, b be OB, and c be OC
The angle between vector a-c and b-c is π/6, which means the angle ACB=30 degrees.
If there is an equilateral triangle OAB, the point C is outside of AB and the angle ACB is a fixed value, it means that the trajectory of C is a circle, and the intersection ACB is the circumferential angle of AB. When C is at the position where OC is the largest, it is obvious that the diameter of C is perpendicular to AB.
And then there's the RT triangle.

Find the coordinates of the unit vector c equal to the included angle of vectors a=(1.2) and b=(2.1

Let C =(cos r,sin r)
Point C multiplied by a and point c multiplied by b are equal
So cos r+2 sin r=2 cos r+sin r
I.e. cos r = sin r apparently
C =(root 2/2, root 2/2) or (-root 2/2,-root 2/2)

The modulus of a vector is known to be 10, b vector to be (3,4), a⊥b Find the coordinates of a

It is readily known that the coordinates of a vector perpendicular to vector b can be expressed as (-4,3)
While vector a⊥ vector b
Let vector a=(-4t,3t)
Given |vector a|=10, then:
|Vector a|2=(-4t)2+(3t)2=10 2
I.e.25 t2=100
Solution: t=2 or -2
So vector a=(-8,6) or a=(8,6)

Given | vector a|=10, vector b=(3,4), and a//b, find the coordinates of vector a

Let a=(x, y)
Because b =(3,4) and a//b
So x/3=y/4 4X=3y (1)
Also|vector a|=10x^2+y^2=10^2=100(2)
The simultaneous (1)(2) solution yields x=6 y=8 or x=-6 y=-8
Therefore, the coordinates of vector a are (6,8) or (-6,8)

Given the absolute value of A=10, B=(3,4), A‖B, then vector A=

Follow vector parallel formula
A (6,8) or (-6,8)

Following vector parallel formula
A (6,8) or (-6,8)

Parallel formula parallel formula
A (6,8) or (-6,8)