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High school compulsory 4 math topics (the following topics are best solved)
1. It is known that non-zero vectors AB and AC satisfy [(AB/| AB |)+(AC/| AC |)]? BC=0,

And (ab/| ab |)? (AC/|AC|) =? , judge the shape of triangle ABC.

(The original title is missing half a bracket) (AB/│AB│ indicates the unit vector in the same direction as the vector AB, with the modulus = 1. The rest are similar)

Solution: (ab/| ab |)? (AC/|AC|)= 1× 1×cosA =? , so a = 60.

[(AB/|AB|)+ (AC/|AC|)]? BC = │( AB/| AB |)+(AC/| AC |)│BC│cos(A/2+C)= 0

Cos(A/2+C)=0, so a/2+c = 90, ∴ c = 90-60/2 = 60.

△ABC is an equilateral△.

2. In quadrilateral ABCD, BD is diagonal, BC=λ(AD)(λ∈R), | AB | = | AD | = 2.

|CB-CD|=2√3

(1), if the triangle BCD is a right triangle, find the value of λ.

(2), under the condition of (1), find CB? barium

Solution: (1) │CB-CD│=│DB│=2√3

In △ABD, │DB│? =│AD│? +│AB│? -2│AD│AB│cosA

That is, 12=4+4-8cosA, so COSA =- 1/2, ∴ A = 120, ∠ Abd = ∠ ADB = 30.

BC=λ(AD), so BC‖AD, and │BC│=λ│AD│=2λ.

∠DBC=∠ABC-∠ABD=60 -30 =30

∠ C = 90, so │ BC │ = 2 (√ 3) COS 30 = 3 = 2 λ, ∴λ = 3/2.

(2)CB? BA =│CB│BA│cos 120 = 3×2×(- 1/2)=-3

3. Make an isosceles right triangle OAB with the origin and point A (5 5,2) as the vertices, so that ∠ b = 90,

Find the coordinates of point b and vector AB.

Solution: │ OA │ = √ 29, the midpoint of OA is M(5/2, 1), with m as the center and │OA│/2=(√29)/2 as the radius to make a circle m:

M: (x-5/2)? +(y- 1)? =29/4

The vertical line intersecting M is OA: Y =-(5/2) (x-5/2)+1=-(5/2) x+29/4, and the equation of inputting M is simplified.

4x? -20x+2 1=(2x-7)(2x-3)=0

Get x? =3.5,x? = 1.5.

So y? =- 1.5,y? =3.5

b? (3.5, - 1.5); b? ( 1.5, 3.5)

Vector AB? =- 1.5i - 3.5j

Vector AB? =-3.5i + 1.5j

4. It is known that the coordinates of O, A and B are O (0 0,0), A (3 3,0) and B (0 0,3) respectively, and the point P is on the line segment AB.

And vector AP=t times vector AB, (0≤t≤ 1), then vector OA? The maximum value of vector OP is _.

Solution: OA? OP =│OA│OP│cos∠AOP≤│OA│? =9

When t=0, that is, point P coincides with point A, OA? OP gets 9 at most.

5, and the vector a=(7/2,? ) and vector b= (? 7/2), the coordinate of the vector with the modulus of 1 is _.

Solution: The unit vector in the same direction as vector a = a/│ a │ = a/(25/2) = 2a/25.

Unit vector b in the same direction as vector b = b/│ b │ = b/(25/2) = 2b/25.

The vector of the sum of a and b is c = a+b = (2/25) (a+b).

Vector c is the square of the frame angle of vectors a and b.

Unit vector c = c/│ c │ = c/(2/25) √ 2 = (a+b)/√ 2 = (√ 2) a/2+(√ 2) b/2.

Therefore, the coordinate of the unit vector c equal to the included angle between vectors A and B is (√2/2, √2/2).

6. Given three points A (1, 2), B (3, 1) and C (- 1, 0), try to answer the following questions:

(1), express vector AB in coordinates, and find its module;

(2) Find the coordinates of point D of vector AB= vector CD;

(3) Let the included angle between vector AB and vector AC be θ, and find the value of cosθ;

(4) Find the area of parallelogram ABCD.

Solution: (1) AB = (3- 1, 1-2) = (2, 1), │ AB │ = √ [2? +(- 1)? ]=√5

(2) Let D(x, y), then CD=(x+ 1, y-0)=(2,-1).

Where x+ 1 = 2, x = 1, y =- 1, so D( 1,-1).

(3)AC=(-2,-2)

What is the slope k of the straight line where AB is located? =- 1/2; What is the slope k of the AC line? = 1

So the tangent tan θ of the included angle θ from AC to AB = (k? -k? )/( 1+k? k? )=(- 1/2- 1)/( 1- 1/2)=-3

So cos θ =-1√ (1+tan? θ)=- 1/√ 10,sinθ= √( 1- 1/ 10)= 3/√ 10,

(4) The area of flat quadrilateral ABCD is s = │ ab │ ac │ sin θ = (√ 5) × (√ 8 )× (3/√10) = 6.

7. In-plane vector OA = (1, 7), vector OB = (5, 1), vector OP = (2, 1), and point Q is a straight line OP.

A moving point on the.

(1), when the vector QA? When the vector QB takes the minimum value, find the coordinates of the vector OQ;

(2) When point Q meets the condition and conclusion of (1), find the value of cos∠AQB.

Solution: (1)Q is on the OP, and let the coordinates of q be (2y, y), where 0≤y≤ 1.

QB=(5-2y, 1-y),QA=( 1-2y,7-y)

QA? QB =(5-2y)( 1-2y)+( 1-y)(7-y)= 5y? -20y+ 12=5(y-2)? -8

QA when y= 1 QB gets the minimum value (-3)

(2) Q (2, 1),QB = (3,0); QA=(- 1,6)

cos∠AQB=QA? QB/│QA│QB │=-3/(3√37)=- 1/√37。

8. Assuming that vectors A and B are non-zero vectors, when the modulus of vector a+t multiplied by vector b (t∈R) is the minimum:

(1), find the value of t;

(2) Verification: Vector B is perpendicular to vector a+t multiplied by vector B. ..

Solution: (1) To simplify the problem, take the intersection o of A and B as the coordinate origin, the vector B is on the X axis, in the same direction as the X axis, and A is in the first quadrant.

The angle between a and b is acute, so a = (m, n), b = (k, 0), (m > 0, n > 0, k > 0).

a+tb=(m+tk,n)

│a+tb│=√[(m+kt)? +n? ]=√(k? t? +2mkt+m? +n? )=√[k? (t+m/k)? +n? ]≥n

The equal sign holds when t=-m/k, where │ a+TB │ min = n and a+TB = (0, n).

(2)b? (a+tb)=k×0+0×n=0, and B? (a+TB)=│b│a+TB│cosθ= 0

Where θ is the included angle between A and a+tb, │b│≠0, │a+tb│≠0, so there must be cosθ=0, that is, θ = 90.

That is, b⊥(a+tb), so the certificate.

9. It is known that AD, BE and CF are the three heights of triangle ABC. Prove that AD, BE and CF intersect at one point.

Solution: It seems difficult to prove this problem with vectors. Look at elementary geometry. Don't ask so many questions at once next time, it's too time-consuming!