![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
Given that the area S of triangle ABC satisfies that the root number 3 is less than or equal to S is less than or equal to 3, and the vector AB×vector BC=6, the included angle between vector AB and vector BC is a,
AB and BC are moduli of vectors AB and BC respectively
Vector AB* Vector BC=AB*BC*cosα=6
S=AB*BC*sin (π-α)/2=AB*BC*sinα/2
√3≤S≤3
3/3≤2S/(vector AB*vector BC)≤1
I.e.√3/3≤tan 1
∴π/6≤α≤π/4
Trigonometric function problem: Let the area of triangle ABC be S, the range of S is from 3 to 3, and the vector AB is multiplied by the vector BC is equal to 6, the angle between vector AB and vector BC is θ. Find:(1) Value range of θ! (2) Find the minimum value of the square of the function f (θ)=(sinθ)+2s in θ cosθ+3 times cosθ. 2. Given the vector m=(cosθ, sinθ) and n=(root number 2-sinθ, cosθ),(π,2π) and |m+n|=5/8 times root number 2, find the value of cos (θ/2/8)?
1
1)
AD vertical BC to D
Area of triangle ABC =1/2* AD*BC =1/2* AB * BC & sinθ
Given that the area S of the triangle ABC satisfies √3≤S≤3, and the vector AB is multiplied by the vector BC equal to 6
(√3)/3≤Sin 1
θ∈[∏/3,2∏/3]
2)
F (θ)=(sinθ)^2+2 s in θ cos 3(cosθ)^2=(sinθ+ cosθ)^2+2(cosθ)^2 2(sinθ+ cosθ)(√2 cosθ)|
The above equation holds only if sinθ+ cosθ=√2 cosθ
When sinθ/cosθ=√2-1.
F (θ)2(sin cosθ)(√2 cosθ)|=(2√2)(tan 1)(cosθ)^2=4(cosθ)^2=4/(1+(tanθ)^2)=4/(4-2√2)=2 2
I.e. when tan θ=√2-1, f (θ) takes the minimum value of 2 2
2.|M||=√(sin 2+ cos 2)=1
| N |=√(2-2√2sin 2+cos 2)=√(3-2√2sin)
|M+n|=(8√2)/5
(1(3-2√2Sinθ))=(8√2)/5
Collate
Sinθ=8/5-(9√2/50)
Cosθ=√(1-sin 2)
Find again
Cos (/4)= cos θ cosπ/4- sin θ sinπ/4
Find again
Cos (θ/2/8)=-√((1+cos (/4))/2)
Given that the area S of triangle ABC satisfies 3≤S≤3*root number 3 and vector AB*vector BC=6, the included angle between vector AB and vector BC is a. Find the minimum value of f (a)= sin^2a+2sinacosa+3cos^2a
|AB|=c;|BC|=a;
3≤S=a*c*sinB/2≤3*Root number 3;(1)
Vector AB* vector BC=6=a*c*cos (180 degrees-B),
So a*c*cosB=-6;(2)
(1)/(2) Simplify:
- Root number 3≤tanB≤-1;
Therefore, the value range of B is:120°≤B≤135°
The calculated angle is the complementary angle of B, so 45°≤a≤60°!
After reduction, f (a)= root 2*sin (2*a+45 degrees)+2(45 degrees≤a≤60 degrees);
Therefore, when a=60°, take the minimum value, and the minimum value is (3+ root number 3)/2;
In a right triangle ABC, where angle C equals 90 degrees and AC equals 4, what is the number product of vector AB and vector AC?
AB*AC=|AB AC|*COS∠A=
(|AB|*COS∠A)*|AC|=
|AC |*|AC |=4*4=16(Tip: The number product of vector AB and AC is equal to the length that AB maps to AC is equal to AC. This length is just AC. You can draw a picture for understanding.)
AB*AC=|AB AC|*COS∠A=
(|AB|*COS∠A)*|AC|=
|AC |*|AC |=4*4=16(Hint: The number product of vector AB and AC is equal to the length of AB mapped to AC and AC is equal to AC. This length is just AC. You can draw a picture for understanding.)
AB*AC=|AB AC|*COS∠A=
(|AB|*COS∠A)*|AC|=
|AC |*|AC |=4*4=16(Hint: The number product of vector AB and AC is equal to the length of AB mapped to AC and AC is equal to the length of AC. This length is just AC. You can draw a picture for understanding.)
In the right triangle ABC, the angle A is 90 degrees, AB=1, and the vector AB is multiplied by the value of the vector BC
Take A as the origin, the line where AB is located is the X axis, and the line where AC is located is the Y axis to establish the rectangular coordinate system
So A (0,0) B (1,0) C (0, y)
So vector AB=(1,0)
Vector BC=(-1, y)
So vector AB point multiplies vector BC=1*(-1)+0*y=-1
Take A as the origin, the line of AB as the X axis, and the line of AC as the Y axis to establish the rectangular coordinate system
So A (0,0) B (1,0) C (0, y)
So vector AB=(1,0)
Vector BC=(-1, y)
So vector AB point multiplies vector BC=1*(-1)+0*y=-1
In right triangle ABC, angle C =90 degrees, AB =5, AC =4, find the product of vector AB and vector BC
|BC|=3, AB*BC=|AB||BC|cos (π-∠B)
=5*3*(-3/5)
= -9
RELATED INFORMATIONS
- 1. In right triangle ABC, if vector A B =(1, k), vector AC =(2,1), and angle C is equal to ninety degrees, then the value of k
- 2. Given vector a =(4.3), vector B =(2,4) then cos
- 3. The condition that a vector plus b vector is perpendicular to a vector minus b vector is A vector =(x1, y1) B vector =(x2, y2)
- 4. In the right triangle ABC, the angle C =90°, AB =5, AC =4 Find the vector AB.BC? Product of quantity! Would you like to start here? In the right triangle ABC, the angle C =90°, AB =5, AC =4 Find the vector A B. BC? Product of quantity! Would you like to start here?
- 5. Given A (-2,4), B (3,-1) C (-3,-4) O as the origin of coordinates, let vector AB = vector a, vector BC = vector b, vector CA = vector c Given A (-2,4), B (3,-1), C (-3,-4). Let vector AB = vector a, vector BC = vector b, vector CA = vector c, and vector CM =3 vector c, vector CN =-2 vector b 1. Find 3a+b-3c 2. Satisfy a=mb+nc
- 6. On-line equal y=a*x^2b*xc AB, BC, CA composition ABC,AB BC CA =0 19.6 X 14.2 x 5.2 cm1n (MN)=1nM 1nM On-line equal y=a*x^2b*xc ternary AB, BC, CA composition ABC,AB BC CA =0 19.6 X 14.2 x 5.2 cm1n (MN)=1nM 1nM
- 7. Product of space vectors Find the volume of a =6,1,-2›, b =0,2,3›, c =6,-2,4› Is it the length of the fork of 3 vectors? Better give me an answer. I'm wrong.
- 8. Let x∈R vector a=(x,1) vector b=(1,-2) and vector a⊥ vector b, then | vector a vector b| equals | Vector a + Vector b | Let x∈R vector a=(x,1) vector b=(1,-2) and vector a⊥ vector b then | vector a vector b| equals | Vector a + Vector b |
- 9. The vector a is known to be equal to (-2.3).b=(5.x), and a⊥b. then x=? A=-15/2 b=-2 The vector a is known to be equal to (-2.3).b=(5.x), and a⊥b. then x=? A=-15/2b=-2/15c=10/3d=10/3 The vector a is known to be equal to (-2.3).b=(5.x) and a⊥b. then x=? A=-15/2 b=-2 The vector a is known to be equal to (-2.3).b=(5.x) and a⊥b. then x=? A=-15/2b=-2/15c=10/3d=10/3 The vector a is known to be equal to (-2.3).b=(5.x), and a⊥b then x=? A=-15/2 b=-2 The vector a is known to be equal to (-2.3).b=(5.x), and a⊥b then x=? A=-15/2b=-2/15c=10/3d=10/3
- 10. A Simple Calculation Using Vector Product → → → → → → → → May I ask why you (a-2b)(2a+b)=2a (^2)-2b (^2)-3a·b (^2) It means square. Thank you, little sister! Using Vector Product to Calculate Simply → → → → → → → → May I ask why the experts (A-2B)(2A+B)=2A (^2)-2B (^2)-3A·B (^2) It means square. Thank you, little sister!
- 11. In triangle ABC, the opposite sides of angles A, B and C are a, b and c respectively, if the product of vector AB and vector AC is equal to the product of vector BA and vector BC and is equal to k (k is a real number) (1) Judge the shape of triangle ABC; (2) If c=√2, find the value of k. In triangle ABC, the opposite sides of angles A, B and C are a, b and c respectively, if the product of vector AB and vector AC is equal to the product of vector BA and vector BC and k (k is real) (1) Judge the shape of triangle ABC; (2) If c=√2, find the value of k.
- 12. Given vector a=(6,4), vector b=(3,8), and x multiplied by vector a+y multiplied by vector b=(-3,5), then x equals y equals Is it done with vector addition?
- 13. If a normal vector of the plane α is n=(4,1,1) and a direction vector of the straight line l is α=(-2,-3,3), then the sine value of the angle between l and α is
- 14. Find the plane equation of M (3,0,-2) with normal vector n=(-2,-4,3)
- 15. Given l, and the direction vector of l is (2, m,1), the normal vector of plane α is (1,1 2,2), Then m=______.
- 16. If the coordinate of vector p at base {abc} is (2010,2011,2012), then the coordinate of vector p at base {a, b+c, b-c} is? If the coordinates of vector p at base {abc} are (2010,2011,2012), then the coordinates of vector p at base {a, b+c, b-c} are?
- 17. In triangle ABC, A=∏/6, and (1+Root 3) c=2b (1) Find angle C (2) If the product of CB vector and CA vector is 1+Root 3, Find a, b, c In triangle ABC, A=∏/6, and (1+Root 3) c=2b (1) Find the angle C (2) If the product of CB vector and CA vector is 1+Root 3, Find a, b, c In triangle ABC, if A=∏/6 and (1+Root 3) c=2b (1), angle C (2) If the product of CB vector and CA vector is 1+Root 3, Find a, b, c
- 18. Vector a=(1,2), b=(1,-1), then /a+2b/=
- 19. Given |a|=2,|b|=1, the angle between a and b is π/3, and the angle between vector 2a+3b and 3a-b is obtained Title Direction sign of omitted vector (→) Given |a|=2,|b|=1, the angle between a and b is π/3, and the angle between vector 2a+3b and 3a-b is obtained Topic Direction sign of omitted vector (→)
- 20. Given the vector a=(3,-1), b=(-1,2), then the coordinate of -3a --2b is? Just the results. Thank you.