Given vector a=(-1, root number three), try to find the unit vector n with an angle of 45 degrees of vector a Given the vector a=(-1, root number three), try to find the unit vector n with an angle of 45 degrees between the sum vector a

Given vector a=(-1, root number three), try to find the unit vector n with an angle of 45 degrees of vector a Given the vector a=(-1, root number three), try to find the unit vector n with an angle of 45 degrees between the sum vector a

Let n=(m, n)
M^2+n^2=1(n is the unit vector)
N·a=‖n a‖cos45(‖n‖ is the absolute value of n)
(M, n)·(-1,√3)=√2
-M 3n=√2
Simultaneous with m^2+n^2=1
Get n=(2√2+√6,√2+√6) or (2√2-√6,-√2+√6)

Let n=(m, n)
M^2+n^2=1(n is the unit vector)
N·a=‖n a‖cos45(‖n‖ is the absolute value of n)
(M, n)·(-1,√3)=√2
-M 3n=√2
Simultaneous with m^2+n^2=1
Get n=(2√2 6,√2 6) or (2√2-√6,-√2 6)

Find the unit vector with an angle of 45 degrees equal to the vector a.

Let b=(x, y), then x^2+y^2=1
Cos=|ab|/(|a||b|)=|(√3-1) x+(√3+1) y|/√8=√2/2
From the above, x=(√6-√2)/4y=(√6 2)/4
Or x=(√6 2)/4 y=(√6-√2)/4

Wait a minute! Find the unit vector whose angle between a =(root 3-1, root 3+1) and vector is 45 degrees.

Let the angle be A and the vector be b (x, y)
CosA =(a*b) divided by the module of the a vector multiplied by the module of the b vector
Therefore, by substituting the number, we can get (root 3-1) x+(root 3+1) y=2
And because it is a unit vector, the square of x + the square of y =1
Therefore, the simultaneous two forms
Get: y =1/2 or y = double root number three
So x = three of two roots or three of two negative roots
When Y = double root number 3,
Because cosA is greater than 0
So x = two-part root number three, y = one-half
So b =(3 of 2,1 of 2)

Given the vector a=(root number 3,1), b is a unit vector that is not parallel to the x-axis, and a*b=root number 3, then b equals Given the vector a=(root number 3,1), b is a unit vector that is not parallel to the x-axis, and a*root number 3, then b equals

Let b=(cosθ, sinθ); from a·b=√3,√3*cosθ+sinθ=√3, then (√3/2)*cosθ+(1/2) sinθ=√3/2, i.e. sin (π/3)*cosθ+cos (π/3)*sinθ=sin (π/3), i.e. sin (/3)=sin (π/3)], i.e./3=2k /2/6; i.e.θ=2kπ or θ=2k...

Given the vector a, b angle is 60 degrees, and | a| is equal to 1,|2a-b| is equal to 2 roots 3. Find | b| The angle of vector a, b is 60 degrees, and |a| equals 1,|2a-b| equals 2. The angle of vectors a, b is known to be 60 degrees, and |a| is equal to 1,|2a-b| is equal to 2. The root number 3.

|2A-b |=2√3
(2A-b)2=12
4|A|2-4|a b cos60 b|2=12
|B|2-2|b|-8=0
Get:
|B |=4

Given that a.b is a unit vector, and |a+2b|=7 under the root sign, then the included angle of vector a.b is equal to why there is only 60 degrees, why not 120 degrees

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