![](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 vector a=(x^2-3,1) b=(x,-y), when x <2 in absolute value, there is a⊥b, when x≥2 in absolute value, a‖b Given vector a=(x^2-3,1) b=(x,-y),(where y and x are not equal to 0), when x <2 in absolute value, there is a⊥b, when x≥2 in absolute value, a‖b, find function y=f (x)
A=(x^2-3,1), b=(x,-y)
Abs (x)=2
(X^2-3)(-y)-x=0
Y=-x%(x^2-3)
So y=(x^2-3) x%2 abs (x)=2.
Note: abs represents absolute value
A=(x^2-3,1), b=(x,-y)
Abs (x)=2
(X^2-3)(-y)-x=0
Y=-x%(x^2-3)
So y =(x^2-3) x%2 abs (x)=2.
Note: abs represents absolute value
Given vector a (-3,4) vector b//a, vector b absolute value is 1, find b, good people have a safe life,
According to the meaning, let vector b be (x, y), so:
Because b//a, there are:
X/y=-3/4.(1)
Because |b|=1
Therefore:
X^2+y^2=1.(2)
Solve the equation group to obtain:
X=-3/5, y=4/5 or x=3/5, y=-4/5.
According to the meaning, let vector b be (x, y), so:
Because b//a, there are:
X/y =-3/4.(1)
Because |b|=1
So:
X^2+y^2=1.(2)
Solve the equation group to obtain:
X=-3/5, y=4/5 or x=3/5, y=-4/5.
According to the meaning, let vector b be (x, y), so:
Because b//a, there are:
X/y =-3/4.(1)
Because |b|=1
Therefore:
X^2+y^2=1.(2)
Solve the equation group to obtain:
X=-3/5, y=4/5 or x=3/5, y=-4/5.
Given that the angle between vector a and vector b is 120°, the absolute value of vector a =3a + the absolute value of vector b = the root number 13, then the absolute value of vector b = Given that the angle between vector a and vector b is 120.d egree. the absolute value of a =3a + the absolute value of b = the root number 13 then the absolute value of b = The absolute value of vector a and b is known to be 120.d egree. a =3 a + b = root 13, then b = absolute value
| A |=3
|A+b|=√13 The square of both sides is:
|^2+2Ab+|b |^2=13
9+2|A||b||cost+|b|^2=13
-3|B 2=4
|^2-3|B|-4=0
(|B|-4)(|b||1)=0
So |b |=4.
| A |=3
|A+b|=√13:
|^2+2Ab||b |^2=13
9+2|A||b||cost+|b|^2=13
-3|B 2=4
|^2-3|B|-4=0
(|B|-4)(|b||1)=0
So |b |=4.
Given vector a parallel vector b, a=(-2,3), b=(1, x+2), find the value of x
Solution
A=(-2,3)
B=(1, x+2)
A//b
A=tb
Then -2=t
3=T (x+2)
Substitute t=-2 into
Then -2(x+2)=3
I.e.-2x-4=3
-2 X =7
X=-7/2
Or directly
Use a=(x1, y1) b (x2, y2)
A//b
X1y2-x2y1=0
If vector a=(1,3), vector b=(x/2,1) and (vector a+2 vector b)⊥2 vector a-vector b) find the value of x If vector a=(1,3), vector b=(x/2,1) and (vector a+2 vector b)⊥2 vector a-vector b
Vector a+2 vector b=(1+x,3+2)=(1+x,5),2 vector a-vector b=(2-x/2,6-1)=(2-x/2,5)
(Vector a+2 vector b)⊥(2 vector a-vector b)
(Vector a+2 vector b)·(2 vector a-vector b)=0, i.e.
(1+X)(2-x/2)+5*5=0.
X^2-3x-54=0
X1=9, x2=-6
Note: If the two vectors are perpendicular, then the dot product (inner product) is equal to 0, and the equation of x is obtained.
Given vector a=(1+cosα, sinα), b=(1-cosβ, sinβ), c=(1,0), a∈(0,π), Let vector a=(1+cosα, sinα), b=(1-cosβ, sinβ), c=(1,0), a∈(0,π),(π,2π), the angle between a and c is θ1, the angle between b and c is θ2, and θ1-θ2=π/6, the value of sinα-β/4 is obtained.
A =(2cos^2(α/2),2sin (α/2) cos (α/2))=2cos (α/2)(cos (α/2), sin (α/2)),
B=(2sin^2(β/2),2sin (β/2) cos (β/2))=2sin (β/2)(sin (β/2), cos (β/2))
Since (0,π),(π,2π),
So α/2∈(0,π/2),β/2∈(π/2,π),
/A/=2cos (α/2),/b/=2s in (β/2)
So cosθ1= cosα/2
So θ1=α/2
Cosθ1= sin (β/2),
Because 0
RELATED INFORMATIONS
- 1. The vector a=(cosα, sinα), b=(cosβ, sinβ),0<β<α If |a-b|=root2 find a perpendicular b let c=(0,1) if a+b=c find the value of The vector a=(cosα, sinα), b=(cosβ, sinβ),0<β<α If |a-b|=rootoot 2, find a perpendicular b Let c=(0,1) If a+b=c, find the value of
- 2. What is the difference between quantity product and vector product? Why the quantity product is perpendicular if the vector a* vector b=0 and the vector a* vector b=0 if the vector product is parallel What is the area of the quantity product and the vector product? Why is the quantity product vertical if the vector a* vector b=0 and the vector product parallel if the vector a* vector b=0?
- 3. Vector a+b=(-2,-1), a-b=(4,-3), the angle between vector a, b is
- 4. Given that M={ a|a=(2 1,-2-2), R}, N={ b|b=(3 2,6 1), R}, is a set of vectors, then M∩N= Options have four A {c|c (-2,1)} B {c|c (2,1)} C {c|c (-2,1)} D {c|c (2,1)}
- 5. Given that point D is the midpoint of the edge BC of the triangle, there is a point P in the plane where the triangle ABC is located, which satisfies vector PA + vector BP + vector CP =0, The module of vector PA is divided by the module of vector PD is equal to λ, and the value of λ is obtained.
- 6. Given vector a=(1, x), vector b=(x^2+x,-x), m is constant and m≤-2 Given vector a=(1, x), vector b=(x^2+x,-x), m is constant and m≤-2 Let a · b +2> m ((2/(a · b))+1). How did x get? Given vector a=(1, x), vector b=(x^2+x,-x), m is constant and m≤-2 Given vector a=(1, x), vector b=(x^2+x,-x), m is constant and m≤-2 Let a · b +2> m ((2/(a · b))+1). How was x found? Given vector a=(1, x), vector b=(x^2+x,-x), m is constant and m≤-2 Given vector a=(1, x), vector b=(x^2+x,-x), m is constant and m≤-2 Let a·b+2> m ((2/(a·b))+1). How did you get x?
- 7. Given vector OA=(3,-4), vector OB=(6,-3), vector OC=(5-x,-3-y), if A, B, C can form a triangle, find the condition that x, y satisfies.
- 8. Point A (-1,0) B (1,4) satisfies vector PA·vector PB=4, and the trajectory equation of point P is obtained. If the point Q is a symmetric point of P with respect to the straight line y=2(x-4), find the trajectory equation of the moving point Q Point A (-1,0) B (1,4) satisfies the vector PA·vector PB=4, and the trajectory equation of point P is obtained. If the point Q is a symmetric point of P with respect to the straight line y=2(x-4), find the trajectory equation of the moving point Q
- 9. In the Cartesian coordinate plane xOy, given the point A (3,2), the point B moves on the circle x^2+y^2=1, the moving point P satisfies the vector AP=vector PB, then the trajectory of the point P is? ... In the Cartesian coordinate plane xOy, given the point A (3,2), the point B moves on the circle x^2+y^2=1, the moving point P satisfies the vector AP=vector PB, then the trajectory of the point P is? Circular elliptic parabolic straight line
- 10. Given the hyperbola 2x2-y2=2, the straight line L passing through the point P (2,1) intersects the hyperbola at two points A and B. If the straight line AB is parallel to the y-axis, the length of the segment AB is obtained. Given the hyperbola 2x2-y2=2, the straight line L passing through the point P (2,1) intersects the hyperbola at the two points A and B. If the straight line AB is parallel to the y-axis, the length of the segment AB is obtained.
- 11. A is n-dimensional orthogonal matrix, a, b is n-dimensional column vector, then Aa·Ab=a·b. Why?
- 12. What is the difference between n-dimensional unit column vector and n-dimensional unit vector
- 13. What does n-dimensional vector mean
- 14. Let the sum of the elements of any row of matrix A of order n be a and prove that a is a eigenvalue of matrix A to find the eigenvector corresponding to a Let a be the sum of the elements of any matrix A of order n be a and prove that a is a eigenvalue of matrix A to find the eigenvector corresponding to a
- 15. Vector midpoint proof In the known isosceles triangle ABC,∠C=90°, D is the midpoint of CB, E is the point on AB, and AE=2EB.
- 16. Is a linear algebra problem," a set of nonzero n-dimensional vectors, if they are orthogonal to each other, then they are called orthogonal vector sets "orthogonal to any two vectors? Linear algebra problem," a set of nonzero n-dimensional vectors, if they are orthogonal to each other, it is called orthogonal vector set "is any two vectors orthogonal? Linear algebra problem," a set of nonzero n-dimensional vectors, if they are orthogonal to each other, then it is called orthogonal vector set "Is any two vectors orthogonal?
- 17. For the vector a, b, c and the real number λ, is the true proposition () in the following propositions possible to analyze these four in detail? A. If a*b=0, then a=0 or b=0 B. If λa=0, then λ=0 or a=0 C. If a^2= b^2, then a = b or a =-b D. If a*b=a*c, then b=c For the vector a, b, c, and real number λ, could the following real proposition () analyze these four in detail? A. If A*B=0, then A=0 or B=0 B. If λa=0, then λ=0 or a=0 C. If A2= B2, then A = B or A = B D. If a*b=a*c, then b=c
- 18. N +1 n-dimensional vector, composed of vector group dimension linear (? ) Vector Group N +1 n-dimensional vector, the vector group composed of linear (? ) Vector Group
- 19. It is proved that in n-dimensional vector space, if α1.α2...αn is linearly independent, then any vector β can be linearly represented by α1.α2..αn. It is proved that any vector β can be linearly represented by α1.α2...α n if α1.α2...
- 20. Let A be the s×m matrix, B be the m×n matrix, and X be the vector of n-dimensional unknown columns. Let A be the s×m matrix, B be the m×n matrix, and X be the vector of n-dimensional unknown columns. AB has the same rank as B, i.e. r (AB)=r (B) Let A be the s×m matrix, B be the m×n matrix, and X be the vector of n-dimensional unknown columns. Let A be the s×m matrix, B be the m×n matrix, and X be the vector of n-dimensional unknown columns. AB has the same rank as B, i.e. r (AB)= r (B)