ABC is known as the sum of three internal angles of triangle ABC, vector m=(-1, root number 3), n=(cosAsinA), and m*n=1. If (1+sin2B)/cos square Bsin square B=3, then tanC ABC is known as the sum of three internal angles of triangle ABC, vector m=(-1, root number 3), n=(cosAsinA), and m*n=1. If (1+sin2B)/cos square Bsin square B=3, find tanC

ABC is known as the sum of three internal angles of triangle ABC, vector m=(-1, root number 3), n=(cosAsinA), and m*n=1. If (1+sin2B)/cos square Bsin square B=3, then tanC ABC is known as the sum of three internal angles of triangle ABC, vector m=(-1, root number 3), n=(cosAsinA), and m*n=1. If (1+sin2B)/cos square Bsin square B=3, find tanC

1. Request A
M·n=(-1)*cosA 3*sinA=√3 sinA-cosA=[(√3)1^]*sin [A-arctan (1/√3)]
=2Sin (A-π/6)
Given m·n=1
Sin (A-π/6)=1/2
A is the inner angle of △ABC
∴0

1. Request A
M·n=(-1)*cosA 3*sinA=√3 sinA-cosA=[(√3)1^]*sin [A-arctan (1/√3)]
=2Sin (A-π/6)
Given m·n=1
Sin (A-π/6)=1/2
A is the internal angle of △ABC
∴0

Given that the included angle of unit vector e1.e2 is 60 degrees, find the included angle of vector a=e1+e2.b=e2-2e1

E1e2=1*1*cos60=1/2
A* b=(e1+e2)(e2-2e1)= e2^2-2e1^1-2e1e2
=1-2-2*1*1*Cos60
=-2
| A||b|= root (e1+e2)^2(e2-2e1)^2
=Root (1+1/2+1)(1-2+2)
=Root (5/2)
So...

E1, e2 are unit vectors, nonzero vector b=xe1+ye2, x, y∈R. If the included angle of e1, e2 is 30°, then (absolute value of x)/(modulus length of b) What is the maximum value for

|E1|=|e2|=1,=π/6
|^2= X^2|e1|^2+y^2||e2|^2+2xye1·e2
=X^2+y^2 3xy
What is the question? Is the maximum value for |xy|/|b|?

|E1|=|e2|=1,=π/6
|^2= X^2|e1|^2+y^2|e2|^2+2xye1·e2
=X^2+y^2 3xy
What is the question? Is the maximum value for |xy|/|b|?

Given that e1 and e2 are unit vectors perpendicular to each other, then e1(e1-e2)= Given that e1 and e2 are unit vectors perpendicular to each other, then e2(e1-e2)=?

E1(e1-e2)= e1*e1-e2*e1
Because they are perpendicular to each other, e1*e2=0
So e1(e1-e2)=e1*e1-e2*e1=e1*e1=1
(Since e1 is a unit vector, e1*e1=1)

Let two non-zero vectors e1 and e2 be not collinear 1. Let m=ke1+e2, n=e1+ke2, and m//n be the value of k Let two nonzero vectors e1 and e2 not be collinear 1. Let m=ke1+e2, n=e1+ke2, and m//n be the value of k

M=ke1+e2, n=e1+ke2, and m//n
So m=in
I.e. ke1+e2=in (e1+ke2)=in e1+in ke2
Therefore:
K = In
1= In k
Get: k=+1 or =-1

Known E1, E2 is two non-zero non-linear vectors, A =2 E1- E2, B=k E1+ E2 if A vs. B is the collinear vector of the number k. Known E1, E2 is two non-zero non-linear vectors, A =2 E1- E2, B=k E1+ E2 if A and B is the collinear vector of the number k. Known E1, E2 is two non-zero non-linear vectors, A =2 E1- E2, B=k E1+ E2, if A and B is the collinear vector of the number k.

A,

B is collinear and there is λ


A =λ

B i.e.2

E1-

E2=λ(k

E1+

E2)
∴
2=λK
-1=λ
K=-2.