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!