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In triangle ABC, the opposite sides of angles A, B and C are a, b and c respectively, and cosC/cosB=3a-c/b, so the value of sinB can be obtained.
It is obtain from that sine theorem,
(3A-c)/b =(3sinA-sinC)/sinB = cosC/cosB
So,
3SinAcosB-sinCcosB=sinBcosC
3SinAcosB=sin (B+C)=sinA
So
CosB =1/3
So
SinB = root number (1-1/9)=(2 root numbers 2)/3
It is obtain from that sine theorem,
(3A-c)/b =(3sinA-sinC)/sinB = cosC/cosB
So,
3SinAcosB-sinCcosB=sinBcosC
3SinAcosB = sin (B+C)= sinA
So
CosB =1/3
So
SinB = root (1-1/9)=(2 root 2)/3
Vector cross multiplication A (vector)×B (vector) Is it the same direction as B (vector)×A (vector)? Are these two formulas equal?
The direction of A (vector)×B (vector) is different from that of B (vector)×A (vector).
Vector cross-multiplication is particular about the order, left and right exchange position, the result is a negative sign;
Point multiplication does not pay attention to the order, left and right exchange position, the result is unchanged.
Given |a|=1,|b|=root2 vector (a-b)⊥a, how much is the angle between a and b?
Because (a-b)⊥a
So (a-b) point is multiplied by the square of a=a-ab=|a|the square of a|-|a||b|cos C tower =1- root number 2cos C tower =0
So cosC tower = root 2/2, C tower between 0 and 90 degrees
So C Tower =45 degrees
Because (a-b)⊥a
So (a-b) point is multiplied by the square of a=a-ab=|a|the square of a|-|a||b|cos C-tower=1-root 2cos C-tower=0
So cosC tower = root 2/2, C tower between 0 and 90 degrees
So C Tower =45 degrees
Given vector a=(2,-2, root number 3), vector b=(-7,0), the angle between vector a and vector b is: Can you write down the steps? Given vector a=(2,-2 n=3), vector b=(-7,0), the angle between vector a and vector b is: Can you write down the steps? Given vector a=(2,-2,3), vector b=(-7,0), the angle between vector a and vector b is: Can you write down the steps?
Let θ be the angle between vector a and vector b
|Vector a|=4
|Vector b|=7
Cosθ=vector a*vector b/|vector a*|vector b|
=-14/28
=-1/2
So θ=120°
Given that A (2,-1), B (-1,1) and O are coordinate origin, dynamic point M satisfies OM=m OA+n OB, where m, n∈R and 2m2-n2=2, then the trajectory equation of M is ______.
Let M be (x, y), then
OM=m
OA+n
OB get
X =2 m−n
Y=m+n, solution obtained
M=x+y
N=x+2y, from 2m2-n2=2, we get x2-2y2=2. So the trajectory equation of point M is x2-2y2=2.
Therefore, the answer is: x2-2y2=2.
Let the coordinates of point M be (x, y), then
OM=m
OA+n
OB get
X =2 m−n
Y=m+n, solution obtained
M=x+y
N=x+2y, from 2m2-n2=2, we get x2-2y2=2. So the trajectory equation of point M is x2-2y2=2.
Therefore, the answer is: x2-2y2=2.
High school mathematics compulsory four vector problem parallel, vertical formula is what? How to calculate the coordinates of the vector? For example, what point is multiplied by what, modulus, modulus of detail You don't have to post that shit. High school mathematics compulsory four vector problem parallel, vertical formula is what? How are the coordinates of the vector calculated? For example, what point is multiplied by what, modulus, modulus of detail You don't have to post that shit.
If vector 1 is (A, B)
Vector 2 is (C, D)
Vector 1.2 is perpendicular to each other.
Then A×C+B×D=0
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