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In the triangle ABC, D is the midpoint of BC. If angle a is equal to 120 degrees and ab vector multiplied by AC vector equals negative 1, then the minimum value of Ad vector modulus is?
Angle a equals 120 degrees
If AB vector * AC vector = - 1, then | ab | * | AC | = 2
Ad vector module = 1 / 2 * radical (| ab | + | AC | - 2)
|Ab | + | AC | ≥ 2 * | ab | * | AC | = 4 (i.e. the minimum value is 4)
The minimum value of Ad vector module = 1 / 2 * root 2
In the triangle ABC, if AB = 2, AC = 3, D is the midpoint of BC edge, then Ad vector multiplies BC vector=
Vector AC - vector AB = vector BC
(vector AC + vector AB) / 2 = vector ad
Ad * BC = (vector AC + vector AB) * (vector AC vector AB) / 2 = (AC square AB Square) / 2 = 2.5
In the triangle ABC, ab = 2, AC = 3, D is the midpoint of BC, then vector ad times vector BC =?
=2.5 firstly, a Cartesian coordinate system is established with a as the origin, AC is set on the X axis, then the trajectory of B is a circle with a as the origin and radius is 2, and the equation is x2 + y2 = 4. Let B be (a, b), then B satisfies the equation. In this case, the coordinates of D are {(3 + a) / 2, B / 2}, vector ad is the same as D coordinate, vector BC is (3-A, - b)
If the angle a = 120 degrees and the vector ab × vector AC = - 2, then the minimum value of | vector Ag | is? The point G is known to be the center of gravity of the triangle ABC AG=1/3(AB+AC) If the angle A=120 degrees, the vector ABX vector AC=-2 Vector ABX vector AC = - 2 = | ab | * | AC | cosa = - 1 / 2 | ab | * | AC| |AB|*|AC|=4 |AG|^2=1/9[|AB|^2+2|AB|*|AC|*cosA+|AC|^2] =1/9[|AB|^2+|AC|^2-|AB|*|AC|] From the mean inequality |AG|^2=1/9[|AB|^2+|AC|^2-|AB|*|AC|]>=1/9(2|AB|*|AC|-|AB|*|AC|)=4/9 AG=2/3 This method solves the problem, but in the beginning, we don't get the value of ab × AC Why can't you start with it AB times AC = 4 Ab ≤ a + B under root 2 of mean inequality Find the minimum value of AB + AC = twice of the center line AG = 2 / 3 midline The answer is different Why not
The logic of the latter method is not very clear The problem lies in the understanding of AB + AC. what you calculate with the mean inequality should be | ab | + | AC |, while the vector Ag = (vector AB + vector AC) / 3, not module addition, may be the problem
How can we get the sentence "the minimum value of AB + AC = twice the central line"
It is known that ad is the median line of Δ ABC, if ∠ a = 120 degrees;, vector AB multiplies vector AC = - 2, then the minimum value of | vector ad | is
A = 120 degrees, vector AB times vector AC = - 2, so | ab | x | AC | 4
2 vector ad = vector AB + vector AC
So 4 | vector ad | 2 = | vector ab | + | vector AC | + 2 vector AB multiplied by vector AC
=|Vector ab| + | vector ac| - 4
≥ 2 | vector ab | vector AC | - 4 = 4
So | vector ad | = 1
So the minimum value of vector ad is 1
In the triangle ABC, vector AB * vector AC = absolute value (vector AB vector AC) = 2. The maximum area of triangle ABC is the size of angle A
Let the three inner angles a, B, C of the triangle ABC be a, B, C, respectively
Then vector ab · vector AC = cbcosa,
Vector AC - vector AB = vector BC,
Because vector ab · vector AC = | vector AC - vector ab | = 3,
So cbcos a = 2, a = 2
According to the cosine theorem, a ^ 2 = B ^ 2 + C ^ 2-2cbcosa,
That is, 4 = B ^ 2 + C ^ 2-4, B ^ 2 + C ^ 2 = 8
So BC ≤ (b ^ 2 + C ^ 2) / 2 = 4,
And BC = 2 / cosa, so 2 / Cosa ≤ 4, Cosa ≥ 1 / 2
So 0
In △ ABC, the vector of 2Ab · the vector of AC = the absolute value of vector ab · the absolute value of vector AC, let ∠ cab = α, ① calculate the value of angle α ② If cos (β - α) = 4 radical 3 / 7, where β ∈ (π / 3,5 π / 6), find the value of cos β
Multiple ifsdoiufjsdcmus
If the nonzero vector AB and AC satisfy (AB / AB + AC / AC) * BC = 0 and AB / absolute value AB * AC / absolute value AC = 0.5, then the triangle ABC is
0
0
Ad is perpendicular to AB, so the vector AB * ad = 0
Vector ac * ad = (AB + BC) * ad
=AB*AD+BC*AD
=0+BC*AD
=√3BD*AD
=√3(BA+AD)*AD
=√3(BA*AD +AD*AD)
=√3(0 +AD*AD)
=√3 AD²
=√3.
In the triangle ABC, the vector ab · vector AC = the square of the absolute value of the vector AC is used to judge the shape of the triangle ABC
Vector ab · vector AC = the square of the absolute value of vector AC
That is vector ab · vector AC = vector AC. vector AC
ν vector ab · vector AC - vector AC · vector AC = 0
ν vector AC · (vector ab - vector AC) = 0
ν vector AC · vector CB = 0
⊥ vector AC ⊥ vector CB
Ψ C is a right angle
The shape of triangle ABC is a right triangle where C is a right angle
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