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How to express A (8.5) B (2.6) vector AB
The coordinate of the vector is equal to the end coordinate minus the start coordinate, i.e.(2-8,6-5)=(-6,1)
Given vector a=(6,8), vector b=(3,4), then |a-b|=explanation process
A-b =(6-3,8-4)=(3,4)
|A-b|=√3 square+4 square=√25=5
Given the constant a >0, the vector c=(0, a), i=(1,0), the straight line passing through the origin O with c+λi as the direction vector and the straight line passing through the fixed point A (0, a) with i-2λc as the direction vector intersect at point P, where R, try to find the trajectory equation of point P
According to the condition of the problem, the equation satisfying the coordinates of the point P is first obtained. According to it is judged whether there are two fixed points, so that the sum of the distances from the point P to the two fixed points is a fixed value. i=(1,0), c=(0, a), c i=(λ, a), i-2λc=(1,-2λa). Therefore, the equations of the straight line OP and AP are λy=ax and y-a=-2λa, respectively.
Given vector m=(2cosX,2sinX), n=(cosX, root number 3 cosX), function f (X)=amn+b-a (a, b is constant and X belongs to R) 1) When a=1, b=2, find out if there is a nonzero integer a, b in the minimum value 2 of f (X), so that when X belongs to [0, p/2], the value range of f (X) is [2,8]. If there is, find out the value of a, b, if not, explain the reason.
Vector m·n=2(cosx)^2+2√3sinxcosx
=1+ Cos2x 3s in 2x
=2(Cos2x*1/2+sin2x 3/2)+1
=2Sin (2x /6)+1,
F (x)=2 asin (2x /6)+a+b-a
=2 Asin (2x /6)+b
When a=1, b=2,
F (x)=2 s in (2x +π/6)+2,
Minimum value:-2+2=0,
When x∈[0,π/2],
(-1/2)2A+b=-a+b,
1*2A+b=2a+b,
F (x) corresponds between -a+b and 2a+b,
-A+b =2,
2A+b =8,
A=2, b=4.
Given the vector m (√3*cosx-sinx,1), n (2cosx, a-√3), x, a∈R, a is a constant.(1) Find y=m*n, y=f (x) Given the vector m (√3*cosx-sinx,1), n (2cosx, a-√3), x, a∈R, a is a constant.(1) Let y=m*n, y=f (x)
Y=m*n=(√3*cosx-sinx)*(2cosx)+1*(a-√3)
=2√3*Cos^(2) x-2sinxcosx+a-√3
=√3(2Cos^(2) x-1)-2sinxcosx+a
=√3 Cos2x-sin 2x + a
= Cos [2x+(π/6)]+a
Y=m*n=(√3*cosx-sinx)*(2cosx)+1*(a-√3)
=2√3*Cos^(2) x-2sinxcosx+a-√3
=√3(2Cos^(2) x-1)-2sinxcosx+a
=√3 Cos2x-sin2x+a
= Cos [2x+(π/6)]+a
Let f (x)=a*b, where a=(m, cosx), b=(1+sinx,1), x∈R, And f (π/2)=2 1 Fact the value of m 2 Find the minimum value of function f (x)
F (x)=m (1+sinx)+cosx=msinx+cosx+m
F (π/2)=2
So m+0+m=2
M=1
So f (x)= sinx + cosx +1= root 2sin (x+45)+1
Then minimum = root 2*(-1)+1=1- root 2
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