In the triangle ABC, if the angle BAC = 120 degrees, ab = 2, AC = 1, D is a point on the edge BC, DC = 2bd, then the vector ad points multiply the vector BC =?

Vector BC = vector AC vector ab
Vector ad = vector AC vector DC
=Vector AC-2 / 3 vector BC
=1 / 3 vector AC-2 / 3 vector ab
Vector ad dot multiplication vector BC = 1 / 3 (vector AC) ^ 2 + 2 / 3 (vector AB) ^ 2 - vector AC dot multiplication vector ab
=1/3+8/3-1*2*COS120
=4

In the triangle ABC, D is the midpoint of BC. If angle a = 120 ° and vector AB is multiplied by AC = - 1, then the minimum value of modulus of ad is

Vector AB times AC = | ab| * | ac* cos120 ° = - 1
|AB|*|AC|=2
Make parallelogram ABEC with AB and AC as adjacent edges,
In △ Abe, ∠ Abe = 60 ° and AE = 2ad
Cosine theorem
AE^2=BA^2+BE^2-2BA*BE*cos60°=BA^2+BE^2-BA*BE>=2BA*BE-BA*BE=BA*BE=2
AE>=√2
AD>=√2/2
The minimum value of the modulus of ad is √ 2 / 2

Given that ad is the median line of the triangle ABC, if the angle a is 120 degrees and the scalar product of vector AB and AC is negative 2, what is the minimum value of vector ad?

According to the central line theorem AB ^ 2 + AC ^ 2 = 1 / 2BC ^ 2 + 2ad ^ 2, i.e. C ^ 2 + B ^ 2 = 1 / 2A ^ 2 + 2ad ^ 2 ∠ a = 120 ° AB * AC = bccos120 ° = - 2, we get BC = 4. Then according to the previous theorem cosa = (b ^ 2 + C ^ 2-A ^ 2) / 2BC = - 1 / 2, we get a ^ 2 = B ^ 2 + C ^ 2 + BC > = 3bC = 12, so ad ^ 2 = 1 / 4A ^ 2-2 {% >

In the triangle ABC, it is known that 2 vector AB city vector AC = absolute value vector AB City absolute value vector AC set angle cab=& Find the value of angle & and if cos (@ ~ &) = 4 / 7, change the sign 3, where @ belongs to the (3 / 5, 5 / 6) find the value of COS @

(1) Cos angle & = 1 / 2, angle - = π / 3
(2) Cos angle & = 1 / 2, sin angle - = √ 3 / 2
cos(@~&) =cos@cos&+sin@sin&=cos@/2 +sin@*√3/2=4√3/7
(sin@)^2+(cos@)^2=1
Combined solution to Cos@

In the triangle ABC, the absolute value of AC vector = 10, the absolute value of Ad vector = 5, Ad vector = 5 / 11db vector, CD vector * AB vector = 0 (1) The absolute value of vector

Fourteen
Ad vector = 5 / 11db vector D on AB ad = 5 BD = 11
CD vector * AB vector = 0 CD is perpendicular to ab
The absolute value of (AB vector AC vector) is the absolute value of CB vector
Pythagorean theorem DC ^ 2 = 75 CB = 14

In triangle ABC, vector ad = 1 / 3 vector AB, vector AE = 1 / 4 vector AC, connected CD, be intersected with P. try to use vector AB and vector AC to represent vector AP

If the parallel line of AB crosses CD at F through point E, then the ratio of EF to ad can be obtained from the ratio of AE to ac. then the ratio of EF to BD can be obtained by knowing the ratio of AD and AB, and then the ratio of EP and Pb can be obtained. Vector EB can be expressed by vector AE and vector AB, so that vector EP can be expressed by AB and AC, and AP = AE + EF. Please complete the calculation by yourself. If you have any questions, please contact me

Let p be a point in △ ABC, and AP=2 Five AB+1 Five AC, then the ratio of △ ABP area to △ ABC area is___ .

Connect CP and lengthen, cross AB to d,
be
AP=2
Five
AB+1
Five
AC=4
Five
AD+1
Five
AC，
Namely
CP=4
PD，
so
CD=5
PD，
Then the ratio of △ ABP area to △ ABC area is 1
5．
So the answer is: 1
Five

In the triangle ABC, am / AB = 1 / 3, an / AC = 1 / 4, BM and cm intersect at point P, and vector AB = a, vector AC = B. try to determine that a and B represent vector AP

Solution 1: let BP = xbn, CP = ycmac + CP = AP = AB + BPAC + y (Ca + AM) = AB + X (Ba + an) B + y (- B + 1 / 3a) = a + X (- A + 1 / 4b) 1-y = x / 41-x = Y / 3x = 8 / 11y = 9 / 11 ﹤ vector AP = AB + BP = a + 8 / 11 (- A + 1 / 4b) = 3 / 11 vector a + 2 / 11 vector B

In the triangle ABC, ad = 2dB, AE = 3ec, CD and be intersect with F, let vector AB = vector a, vector AC = vector B, vector AF = x * vector a + y * vector B What is (x, y) The answer is (1 / 3,1 / 2)

If the vectors AB = a, ad = 2 / 3 * a, AC = B, AE = 3 / 4 * B, CD = ad, AC = 2 / 3 * A-B, be = ae-ab = 3 / 4 * B-A, C, D, D, f are collinear, then CF = MCD = 2m / 3 * A-M * AF = AC + CF = 2m / 3 * A-M * AF = AC + CF = 2m = 3 * 3 * a * a + (1-m) * B (1) B, e, f are collinear, then BF = NBE = 3N / 4 * B-N * n * AF = AF = AB + BF = 3N / 4 * B + 1-N * B (1-N) * a (2) a (2) a (2) a (2) a (2) B = 3 = 3 * 4 * 4 and base

In the triangle ABC, am: ab = 1:3, an: AC = 1:4, BN and cm intersect at p. if AB vector = a vector, AC vector = B vector, find AP vector,

Let BP = xbn, CP = YCM
AC+CP=AP=AB+BP
AC+y(CA+AM)=AB+x（BA+AN)
b+y(-b+1/3a)=a+x(-a+1/4b)
1-y=x/4
1-x=y/3
x=8/11
y=9/11
The vector AP = AB + BP = a + 8 / 11 (- A + 1 / 4b) = 3 / 11 vector a + 2 / 11 vector B

In the triangle ABC, if the angle BAC = 120 degrees, ab = 2, AC = 1, D is a point on the edge BC, DC = 2bd, then the vector ad points multiply the vector BC =?