Given vector a=(2,1), vector b=(-8,6) (1) Given vector a=(-1,2), vector b=(1,-2), find the coordinates of vector a+vector b, vector a-vector b,2 vector a-3 vector b. (2) Given vector a=(2,1), vector b=(-8,6), vector c=(4,6), find 2 vector a+5 vector b-vector c, and express the vector with base vector i, j

Given vector a=(2,1), vector b=(-8,6) (1) Given vector a=(-1,2), vector b=(1,-2), find the coordinates of vector a+vector b, vector a-vector b,2 vector a-3 vector b. (2) Given vector a=(2,1), vector b=(-8,6), vector c=(4,6), find 2 vector a+5 vector b-vector c, and express the vector with base vector i, j

(1) Given vector a=(-1,2), vector b=(1,-2), find the coordinates of vector a+vector b, vector a-vector b,2 vector a-3 vector b.
(2) Given vector a=(2,1), vector b=(-8,6), vector c=(4,6), find 2 vector a+5 vector b-vector c, and express the vector with base vector i, j

Given constant a >0, passing through fixed point A (0, a), a line with m vector =(λ, a) as direction vector and passing through fixed point B (0, a), and n vector =(1,2λa)... Is the line of the direction vector intersects P, where λ is R, and find the equation of the trajectory C of point P Given constant a >0, passing through fixed point A (0, a), a line with m vector =(λ, a) as direction vector and passing through fixed point B (0, a), and n vector =(1,2λa)... Is the straight line of the direction vector intersects at P, where λ is R, and find the equation of the locus C of the point P

Let P (x, y), vector AP=(x, y+a), vector BP=(a, y-a)
Vector m is the direction vector vector m// vector AP ax=λ(y+a), the same as 2λax=y-aλ=(y-a)/2ax
Eliminate λ ax=(y+a)*(y-a)/2ax y^2/a^2-2x^2=1 It is hyperbolic

Let a=(1, e^-x), b=(e^x, m), where m is a constant and m∈R. When m=-1, find the inequality f (x^2-3)+f (2x)

When m=-1, f (x)=e^x-e^(-x),
F'(x)=e^(x)+e^(-x)>0
Then f (x)=e^x-e^(-x), is the increasing function on R.
F (x^2-3)+f (2x)

When m=-1, f (x)=e^x-e^(-x),
F'(x)=e^(x)+e^(-x)>0
Then f (x)=e^x-e^(-x), is the increasing function on R.
Then f (x^2-3)+f (2x)

Given a is constant and a >0, vector m=(√x,-1), vector n=(1, ln (x+a)), Find the Maximum Value of Function f (x)=m·n in Interval (0,1]

F (x)=m•n=√x-ln (x+a) f'(x)=1/(2*(x))-1/(x+a)=(x+a-2(x))/(2(x)*(x+a)) According to the condition x >0, a >0, so the denominator is greater than 0, only the numerator needs to be observed.

Given vector a=(m, n), b=(coswx,sinwx), where m, n, w are constants, and w >0, x∈R, the function y=f (x)=vector a*vector b has a period of π, when x=π,12, the function obtains a maximum value of 1. (1) Find the analytic expression of function f (x) (2) Write the axis of symmetry of y = f (x) and prove it Given vector a=(m, n), b=(coswx,sinwx), where m, n, w are constants, and w >0, x∈R, the function y=f (x)=vector a*vector b has a period of π, when x=π,12, the function obtains a maximum value of 1. (1) Analytic expression of function f (x) (2) Write the axis of symmetry of y=f (x) and prove it

Given vector a=(m, n), b=(coswx,sinwx), where m, n, w are constants, and w >0, x∈R, the function y=f (x)=vector a*vector b has a period of π, when x=π/12, the function obtains a maximum value of 1.
(1) Find the analytic expression of function f (x)
(2) Write the axis of symmetry of y = f (x) and prove it
[Solution]: vector a* vector b=mcoswx+nsin wx=√(m^2+n^2) sin (wx )[ tanφ=m/n]
Period is π,==> w=2
1 Max when x =π/12
√(M^2+n^2)=1
π/6=π/2+2Kπ
Solution m=3/2, n=±1/2
So f (x)= sin (2x /3).
[2]: Axis of symmetry 2x /3=π/2+kπ
Solution: x=π/12+kπ/2

Given three-point space A (0,2,3) B (-2,1,6) C (1,-1,5). Given three points A (0,2,3) B (-2,1,6) C (1,-1,5).

Vector AB=(-2,-1,3); AC=(1,-3,2); BC=(3,-2,1)
It is found that modulus of AB=module of AC=module of BC=√14;
ABC is regular triangle,∠A=60°
Height of parallelogram=AB*sinA=(√42)/2;
Area of parallelogram=bottom*height=AB*(√42)/2=7√3