On the scalar product of plane vectors Verification: the three high Lines ad, be, CF of △ ABC intersect with a point. (please use the idea of plane vector to prove)

Let Δ ABC, the three high lines are ad, be, cf. AD and be intersect with H and connect cf. vector ha = vector a, vector HB = vector B, vector HC = vector C
Because ad ⊥ BC, be ⊥ AC,
So vector ha · vector BC = 0, vector HB · vector CA = 0,
That is, vector a · (vector c-vector b) = 0,
Vector B · (vector a-vector C) = 0,
or
Vector a · vector C - vector a · vector b = 0
Vector B · vector a - vector B · vector C = 0
The two formulas are added together
Vector C · (vector a-vector b) = 0
The vector HC · vector Ba = 0
Therefore, CH ⊥ AB, C, F, h are collinear, ad, be, CF intersect at the same point, H

Several problems about the scalar product of plane vector 1. Given that a and B are nonzero vectors, when t =? The modulus of a + TB is the minimum 2. Set 0

When t = 0, the sum of the two sides of the triangle is greater than the third side. A, TB, a + TB form a vector triangle. When t = 0, the module of a + TB is the smallest of A
2.y=-sin^2 x-│a│sinx-│b│+1=-(sinx+│a│/2)^2-│b│+1+│a│^2/4
-Therefore, when SiNx = 1, y is minimum and Y is maximum at the apex of parabola
The minimum value of Y is - a │ - B │, and the maximum value of Y is - │ B │ + 1 + │ a │ ^ 2 / 4
So, B + a = 4
-│b│+1+│a│^2/4=0
The results show that │ B │ = 2 │ a │ = 2. In addition, a │ = 6 > 2 is omitted
a. B, a + B form a diamond, vector (a + b) ^ 2 = a ^ 2 + B ^ 2 + 2 │ B │ a │ cosa = 4 + 4 + 2 * 4 √ 2 / 2 = 8 + 4 √ 2
│a+b│=2√(2+√2)

The scalar product of a plane vector! Given the vector a = (2, λ), vector b = (3, - 4), the angle between tangent vector a and B is obtuse angle, then the value range of λ___ Given that vector a = (2,3) | vector B | = √ 13, vector a ∥ vector B, then the coordinates of vector B_ Another question is, what's the difference between vector a and vector a?

Solution: cos α = AB / | a | B | 0
(2,3) (- 2, - 3) solution: ∵ a ∥ B a = λ B 2 = λ x 3 = λ y √ x2 + y2 = √ 13  λ = 1, - 1
a=(x,y) |a|=√x2+y2

A complete set of high school mathematics formulas

I have a copy of all the formulas, including physical chemistry
Parabola: y = ax * + BX + C
That's y equals the square of ax plus BX plus C
When a > 0, the opening is upward
When a < 0, the opening is downward
When C = 0, the parabola passes through the origin
When B = 0, the parabola symmetry axis is y-axis
And the vertex y = a (x + H) * + K
That's y equals a times the square of (x + H) + K
-H is x of the vertex coordinates
K is y of the vertex coordinates
It is generally used to find the maximum and minimum values
Parabolic standard equation: y ^ 2 = 2px
It means that the focus of the parabola is on the positive half axis of X, the focus coordinate is (P / 2,0), and the directrix equation is x = - P / 2
Since the focus of the parabola can be on any half axis, the common standard equation y ^ 2 = 2px y ^ 2 = - 2px x x ^ 2 = 2PY x ^ 2 = - 2PY x ^ 2 = - 2PY
Circle: volume = 4 / 3 (PI) (R ^ 3)
Area = (PI) (R ^ 2)
Perimeter = 2 (PI) r
The standard equation of a circle is (x-a) 2 + (y-b) 2 = R2. Note: (a, b) is the center coordinate of the circle
The general equation of circle x2 + Y2 + DX + ey + F = 0 note: D2 + e2-4f > 0
（1） Calculation formula of ellipse circumference
Ellipse perimeter formula: l = 2 π B + 4 (a-b)
The circumference theorem of ellipse: the circumference of an ellipse is equal to the difference between the circumference of the circle whose short half axis length is the radius (2 π b) plus four times the length of the ellipse's half axis (a) and the short half axis length (b)
（2） Calculation formula of ellipse area
Ellipse area formula: S = π ab
The area theorem of ellipse: the area of an ellipse is equal to the product of the circumference (π) times the length of the long half axis (a) and the length of the short half axis (b) of the ellipse
Although there is no ellipticity t in the above formula of circumference and area of ellipse, these two formulas are derived from the derivation of ellipticity T. the constant is a volume, and the formula is used
Formula for calculating the volume of an ellipse the long radius * the short radius * the PAI * height of the ellipse
Trigonometric function:
Sum of two angles formula
sin(A+B)=sinAcosB+cosAsinB sin(A-B)=sinAcosB-sinBcosA
cos(A+B)=cosAcosB-sinAsinB cos(A-B)=cosAcosB+sinAsinB
tan(A+B)=(tanA+tanB)/(1-tanAtanB) tan(A-B)=(tanA-tanB)/(1+tanAtanB)
cot(A+B)=(cotAcotB-1)/(cotB+cotA) cot(A-B)=(cotAcotB+1)/(cotB-cotA)
Double angle formula
tan2A=2tanA/(1-tan2A) cot2A=(cot2A-1)/2cota
cos2a=cos2a-sin2a=2cos2a-1=1-2sin2a
sinα+sin(α+2π/n)+sin(α+2π*2/n)+sin(α+2π*3/n)+…… +sin[α+2π*(n-1)/n]=0
cosα+cos(α+2π/n)+cos(α+2π*2/n)+cos(α+2π*3/n)+…… +Cos [α + 2 π * (n-1) / N] = 0 and
sin^2(α)+sin^2(α-2π/3)+sin^2(α+2π/3)=3/2
tanAtanBtan(A+B)+tanA+tanB-tan(A+B)=0
Four fold angle formula:
sin4A=-4*(cosA*sinA*(2*sinA^2-1))
cos4A=1+(-8*cosA^2+8*cosA^4)
tan4A=(4*tanA-4*tanA^3)/(1-6*tanA^2+tanA^4)
Five fold angle formula:
sin5A=16sinA^5-20sinA^3+5sinA
cos5A=16cosA^5-20cosA^3+5cosA
tan5A=tanA*(5-10*tanA^2+tanA^4)/(1-10*tanA^2+5*tanA^4)
Hexagonal formula:
sin6A=2*(cosA*sinA*(2*sinA+1)*(2*sinA-1)*(-3+4*sinA^2))
cos6A=((-1+2*cosA^2)*(16*cosA^4-16*cosA^2+1))
tan6A=(-6*tanA+20*tanA^3-6*tanA^5)/(-1+15*tanA^2-15*tanA^4+tanA^6)
Seven fold angle formula:
sin7A=-(sinA*(56*sinA^2-112*sinA^4-7+64*sinA^6))
cos7A=(cosA*(56*cosA^2-112*cosA^4+64*cosA^6-7))
tan7A=tanA*(-7+35*tanA^2-21*tanA^4+tanA^6)/(-1+21*tanA^2-35*tanA^4+7*tanA^6)
Octagonal formula:
sin8A=-8*(cosA*sinA*(2*sinA^2-1)*(-8*sinA^2+8*sinA^4+1))
cos8A=1+(160*cosA^4-256*cosA^6+128*cosA^8-32*cosA^2)
tan8A=-8*tanA*(-1+7*tanA^2-7*tanA^4+tanA^6)/(1-28*tanA^2+70*tanA^4-28*tanA^6+tanA^8)
Nine fold angle formula:
sin9A=(sinA*(-3+4*sinA^2)*(64*sinA^6-96*sinA^4+36*sinA^2-3))
cos9A=(cosA*(-3+4*cosA^2)*(64*cosA^6-96*cosA^4+36*cosA^2-3))
tan9A=tanA*(9-84*tanA^2+126*tanA^4-36*tanA^6+tanA^8)/(1-36*tanA^2+126*tanA^4-84*tanA^6+9*tanA^8)
Ten fold angle formula:
sin10A=2*(cosA*sinA*(4*sinA^2+2*sinA-1)*(4*sinA^2-2*sinA-1)*(-20*sinA^2+5+16*sinA^4))
cos10A=((-1+2*cosA^2)*(256*cosA^8-512*cosA^6+304*cosA^4-48*cosA^2+1))
tan10A=-2*tanA*(5-60*tanA^2+126*tanA^4-60*tanA^6+5*tanA^8)/(-1+45*tanA^2-210*tanA^4+210*tanA^6-45*tanA^8+tanA^10)
·Universal formula:
sinα=2tan(α/2)/[1+tan^2(α/2)]
cosα=[1-tan^2(α/2)]/[1+tan^2(α/2)]
tanα=2tan(α/2)/[1-tan^2(α/2)]
Half angle formula
sin(A/2)=√((1-cosA)/2) sin(A/2)=-√((1-cosA)/2)
cos(A/2)=√((1+cosA)/2) cos(A/2)=-√((1+cosA)/2)
tan(A/2)=√((1-cosA)/((1+cosA)) tan(A/2)=-√((1-cosA)/((1+cosA))
cot(A/2)=√((1+cosA)/((1-cosA)) cot(A/2)=-√((1+cosA)/((1-cosA))
Sum difference product
2sinAcosB=sin(A+B)+sin(A-B) 2cosAsinB=sin(A+B)-sin(A-B)
2cosAcosB=cos(A+B)-sin(A-B) -2sinAsinB=cos(A+B)-cos(A-B)
sinA+sinB=2sin((A+B)/2)cos((A-B)/2 cosA+cosB=2cos((A+B)/2)sin((A-B)/2)
tanA+tanB=sin(A+B)/cosAcosB tanA-tanB=sin(A-B)/cosAcosB
cotA+cotBsin(A+B)/sinAsinB -cotA+cotBsin(A+B)/sinAsinB
Sum of the first n terms of some sequences
1+2+3+4+5+6+7+8+9+… +n=n(n+1)/2 1+3+5+7+9+11+13+15+… +(2n-1)=n2
2+4+6+8+10+12+14+… +(2n)=n(n+1) 1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+… +n^2=n(n+1)(2n+1)/6
1^3+2^3+3^3+4^3+5^3+6^3+… n^3=(n(n+1)/2)^2 1*2+2*3+3*4+4*5+5*6+6*7+… +n(n+1)=n(n+1)(n+2)/3
Sine theorem a / Sina = B / SINB = C / sinc = 2R note: where R is the circumscribed radius of a triangle
Cosine theorem B2 = A2 + c2-2accosb note: angle B is the angle between edge a and edge C
Multiplication and factorization A2-B2 = (a + b) (a-b) A3 + B3 = (a + b) (a2-ab + B2) a3-b3 = (a-b (A2 + AB + B2)
Trigonometric inequality | a + B | | a | + | B | A-B | ≤| a | + | B | a | ≤ B-B ≤ a ≤ B
|a-b|≥|a|-|b| -|a|≤a≤|a|
Solutions of quadratic equation of one variable - B + √ (b2-4ac) / 2A - B - √ (b2-4ac) / 2A
The relation between root and coefficient X1 + x2 = - B / a X1 * x2 = C / a note: Weida theorem
The discriminant b2-4a = 0 note: the equation has two equal real roots
B2-4ac > 0 note: the equation has two unequal real roots
b2-4ac0
Parabolic standard equation y2 = 2px y2 = - 2px x2 = 2PY x2 = - 2PY x2 = - 2PY
Side area of straight prism s = C * h, side area of oblique prism s = C '* h
The side area of regular pyramid s = 1 / 2C * h'the side area of regular pyramid s = 1 / 2 (c + C ') H'
The side area of the cone s = 1 / 2 (c + C ') l = pi (R + R) l the surface area of the ball s = 4Pi * R2
Cylinder side area s = C * H = 2pi * h, conical side area s = 1 / 2 * c * l = pi * r * l
The arc length formula L = a * r a is the radian number of the central angle R > 0, and the fan area formula s = 1 / 2 * L * r
Cone volume formula v = 1 / 3 * s * h cone volume formula v = 1 / 3 * pi * r2h
Volume of oblique prism v = s'L note: where s' is the area of straight section and l is the length of side edge
Volume formula of cylinder v = s * h cylinder v = pi * r2h
Volume formula of perimeter area of figure
Perimeter of rectangle = (length + width) × 2
Perimeter of square = side length × 4
Area of rectangle = length × width
Area of square = side length × side length
The area of a triangle
Given the triangle base a, height h, then s = ah / 2
If a, B, C and half circumference P of a triangle are known, then s = √ [P (P-A) (P-B) (P-C)] (Helen's formula) (P = (a + B + C) / 2)
And: (a + B + C) * (a + B-C) * 1 / 4
Given a, B on both sides of the triangle, and the angle c between these two sides, then s = absinc / 2
Let the three sides of the triangle be a, B and C respectively, and the radius of the inscribed circle is r
Then the triangle area = (a + B + C) R / 2
Let the three sides of the triangle be a, B and C respectively, and the radius of the circumscribed circle is r
Then the triangle area = ABC / 4R
If we know the triangles a, B and C, then s = √ {1 / 4 [C ^ 2A ^ 2 - ((C ^ 2 + A ^ 2-B ^ 2) / 2) ^ 2]}
| a b 1 |
S△=1/2 * | c d 1 |
| e f 1 |
【| a b 1 |
|C D 1 | is a third-order determinant. This triangle ABC is in plane rectangular coordinate system a (a, b), B (C, d), C (E, f), where ABC
| e f 1 |
It's better to start from the upper right corner in anti clockwise order, because the results obtained in this way are generally positive. If you don't follow this rule, you may get a negative value, but it doesn't matter. As long as you take the absolute value, it won't affect the size of the triangle area! "
The area formula of the middle line of qinjiushao triangle is as follows:
S=√[(Ma+Mb+Mc)*(Mb+Mc-Ma)*(Mc+Ma-Mb)*(Ma+Mb-Mc)]/3
Where Ma, MB and MC are the midline lengths of the triangle
Area of parallelogram = base × height
Area of trapezoid = (upper bottom + bottom) × height △ 2
Diameter = radius × 2 radius = diameter △ 2
Circumference of circle = Pi × diameter=
Circumference × radius × 2
The area of a circle = circumference × radius × radius
Surface area of cuboid=
(length × width + length × height + width × height) × 2
Volume of cuboid = length × width × height
The surface area of cube = edge length × edge length × 6
The volume of cube = edge length × edge length × edge length
The side area of the cylinder = perimeter × height of the bottom circle
The surface area of the cylinder = upper and lower bottom surface area + side area
Volume of cylinder = base area × height
Volume of cone = base area × height △ 3
Cuboid (cube, cylinder)
Volume = bottom area × height
Plane figure
Name symbol perimeter C and area s
Square a - side length C = 4A
S＝a2
Rectangle A and B - side length C = 2 (a + b)
S＝ab
Triangle a, B, C - length of three sides
The height on H-A edge
S - half of perimeter
A. B, C - internal angle
Where s = (a + B + C) / 2 s = ah / 2
＝ab/2?sinC
＝[s(s-a)(s-b)(s-c)]1/2
＝a2sinBsinC/(2sinA)
There is and only one straight line after two
2. The line segment between two points is the shortest
The complementary angles of the same angle or equal angle are equal
The remainder of the same angle or equal angle is equal
There is and only one line perpendicular to the known line
Among all the lines connected by a point outside the line and the points on the line, the vertical line is the shortest
The parallel axiom passes through a point outside the line, and there is only one line parallel to this line
If both lines are parallel to the third line, they are parallel to each other
The two lines are parallel
The internal staggered angle is equal and the two straight lines are parallel
The inner angle of the same side is complementary, and the two straight lines are parallel
The two lines are parallel and the same angle
The two straight lines are parallel and the angle of internal stagger is equal
The two straight lines are parallel and complementary to each other
Theorem 15 the sum of the two sides of a triangle is greater than the third side
16 infer that the difference between the two sides of the triangle is less than the third side
The sum of the three interior angles of a triangle is equal to 180 degrees
Inference 1 two acute angles of a right triangle are complementary
Inference 2 an outer angle of a triangle is equal to the sum of two interior angles that are not adjacent to it
Inference 3 an outer angle of a triangle is greater than any inner angle that is not adjacent to it
The corresponding sides and angles of an congruent triangle are equal
There are two congruent triangles corresponding to two sides and their included angles
There are two congruent triangles corresponding to two angles and their sandwich edges
Inference (AAS) there are two triangles congruent with two angles and the opposite side of one of them
The 25 side side side axiom (SSS) has three sides corresponding to the congruence of two triangles
The axiom of hypotenuse and right angled side (HL) two right triangle congruences with a hypotenuse and a right angle side corresponding to each other
Theorem 1 the distance from a point on the bisector of an angle to both sides of the angle is equal
Theorem 2 a point with the same distance from both sides of an angle is on the bisector of the angle
The bisector of an angle 29 is the collection of all points equidistant from both sides of the angle
Property theorem of isosceles triangle two base angles of isosceles triangle are equal
Corollary 1 the bisector of the vertex angle of an isosceles triangle bisects and is perpendicular to the base
The bisector of the top angle, the center line on the bottom edge and the height on the bottom edge of an isosceles triangle coincide with each other
Corollary 3 the angles of an equilateral triangle are equal, and each angle is equal to 60 degrees
Theorem of isosceles triangle if two angles of a triangle are equal, the opposite sides of the two angles are equal (equal angle to equal side)
Corollary 1 a triangle whose three angles are equal is an equilateral triangle
Corollary 2 an isosceles triangle with an angle equal to 60 ° is an equilateral triangle
In a right triangle, if an acute angle is equal to 30 degrees, the right angle it faces is equal to half of the hypotenuse
The center line on the hypotenuse of a right triangle is equal to half of the hypotenuse
Theorem 39 the distance between the point on the vertical bisector of a line segment and the two endpoints of the line segment is equal
40 inverse theorem and a line segment two endpoints of the same distance, in the vertical direction of this line segment

Coordinate operation of plane vector Given the point (2,3), (5,4), (7,10) if (vector) AP = AB + MAC (M belongs to R), when trying to find what value m is, 1. Point P is on the bisector of the first and third quadrant angle; 2. Point P is in the third quadrant!

Point a (2,3), B (5,4), C (7,10)
Then the vector AB = (3,1), AC = (5,7)
AP=AB+mAC=(3+5m,1+7m)
1. Point P is on the bisector of the first and third quadrants;
Then there is: y = X
That is, 3 + 5m = 1 + 7m
M=1
2. Point P is in the third quadrant
Then there are: x < 0, y < 0
That is: 3 + 5m < 0, 1 + 7m < 0
m<-3/5,m<-1/7
That is, m < - 3 / 5

The plane coordinates of this problem A (- 2,4), B (3, - 1), C (- 3,4) are known Let AB = a, BC = B, CA = C Find 3A + B - 3C; This problem is solved as follows: ∵ a (- 2,4), B (3, - 1), C (- 3,4) ∴ a = AB = （5,-5） b = BC = (-6,-3) C = CA = (1,8) How to calculate the last three coordinates? Is the end coordinate minus the starting point coordinate? How do I calculate and the answer is not the same?

The calculation result of a is correct, B = (- 6,5), C = (- 1,0). It is estimated that there is a negative sign in the wrong position. The vector calculation method of plane rectangular coordinate system is that the end coordinate minus the starting point coordinate

Coordinate operation of plane vector? Given the coordinates of vectors a and B, find a + B, a-b, 1.a=(-2,4),b=(5,2) 2.a=(4,3),b=(-3,8) 3.a=(2,3),b=(-2,-3) 4.a=(3,0),b=(0,4) Let a = (3,2) B = (0, - 1) find the coordinates of - 2A + 4b, 4A + 3B,

1 a+b=(-2+5,4+2)=(3,6)a-b=(-2-5,4-2)=(-7,2)2 a+b=(4+(-3),3+8)=(1,11)a-b=(4-(-3),3-8)=(7,-5)3 a+b=(2+(-2),3+(-3))=(0,0)a-b=(2-(-2),3-(-3))=(2,6)4 a+b=(3+0,0+4)=(3,4)a-b=(3-0,0-4)=(3,-4)-2a+4b=-2(3,2)+4...

On the coordinate operation of plane vector A vector corresponds to a unique coordinate Equal vector coordinates are the same A coordinate corresponds to a unique vector A point in the plane corresponds to a vector starting from the origin and ending at that point The right and wrong of these four propositions and their causes, Why?

A pair, each vector has its own unique coordinate representation B pair, the vector is equal to their coordinates, the same in a and B, it should be noted that a coordinate can represent countless vectors, such as the vector whose starting point is (1,1), the ending point is (2,3), and the vector whose starting point is (0,0), and the vector whose ending point is (1,2), their coordinate representation is (1,2

What is the operation rule of parallelogram in plane vector operation?... it is better to be formula type

When two forces are combined, a parallelogram is made by taking the line segment representing the two forces as the adjacent side. The diagonal line between the two adjacent sides represents the magnitude and direction of the resultant force. This is called the parallelogram rule

Plane vector multiplication Vector a = (x1, Y1) vector b = (X2, Y2) So what is the multiplication of a and B in coordinates?

x1x2+y1y2

On the scalar product of plane vectors Verification: the three high Lines ad, be, CF of △ ABC intersect with a point. (please use the idea of plane vector to prove)