In triangle ABC vector A B =4 vector AC =2 vector AD =1/3 vector AB +2/3 vector AC proves BCD three points collinear when vector AD = root 6 find vector BC In triangle ABC vector A B =4 vector AC =2 vector AD =1/3 vector AB +2/3 vector AC proves BCD three point collinear when vector AD = root 6 find vector BC

In triangle ABC vector A B =4 vector AC =2 vector AD =1/3 vector AB +2/3 vector AC proves BCD three points collinear when vector AD = root 6 find vector BC In triangle ABC vector A B =4 vector AC =2 vector AD =1/3 vector AB +2/3 vector AC proves BCD three point collinear when vector AD = root 6 find vector BC

BD=AD-AB=(1/3) AB+(2/3) AC-AB=(2/3) AC-(2/3) AB=(2/3)(AC-AB) BC=AC-AB (2/3) BC=BD, so BC//BC and B are common points, so BCD is collinear|(1/3) AB|=4/3|(2/3) AC|=4/3AD is an angular branch of diamond with side length of 4/3. cos (A/2)=3√6/8cosA=2cos (A/2)^2-1=1...

In △ABC, there is a point D on BC, and if the vector AD =2/3 vector AB +1/3 vector AC, then () A.∠BAD∠CAD C.|Vector BD|>|Vector DC| D.|VECTOR BD|

Point D is the quarter of BC.
You don't have to choose DA. You never choose math.
Senior senior's advice: generally no special instructions are not multiple choice

In △ABC, AB • AC =−1 3, AB • BC =1. Find:(1) length of AB side; (2) Find sin (A−B) 3SinC.

(1)

AB •

AC =

AB•(

AB+

BC)=

AB2+

AB •

BC
=

AB 2-3=1
∴|

AB|2=4,|

AB |=2...(5 points)
(2) From (1),2bcosA=1,2acosB=3
3BcosA=acosB
:3SinBcosA by sine theorem = sinAcosB...(8 points)
Sin (A-B)
3SinC = sinAcosB-sinBcosA
3(SinAcosB+sinBcosA)=1
6...(12 Points)

In triangle ABC, DE parallel BC intersects AB, AC at D, E, if vector AD =1/3AB, vector AB = vector a, vector AC = vector b, then vector DE =? The answer is 1/3(vector b-vector a) Why?

△ADE is similar to △ABC,|AD|/|AB|=|AE|/|AC|=|DE|/|BC|=1/3
And vector BC=AC-AB=b-a, so vector DE=BC/3=(b-a)/3

O is a fixed point on the plane, A, B, C are three non-collinear points on the plane, and the moving point P satisfies the vector OP=OA+t (AB+AC), t∈[0,+∞). Then the trajectory of P Must pass through the heart of △ABC

Let the midpoint of BC be D, then the trajectory of p is the straight line of AD, and then what heart, forget, is the intersection of the three center lines.

If the non-zero vectors A and B are in the same or opposite direction, must vector A plus vector B be in the same direction as vector A. Why? If the non-zero vectors A and B are in the same or opposite direction, must vector A plus vector B be in the same direction as vector A and vector B? Why? Why? If the non-zero vectors A and B have the same or opposite directions, must vector A add vector B in the same direction as vector A. Why?

No. No.
(1) When the directions of the non-zero vectors A and B are the same, the direction of the vector A+vector B is the same as the direction of the vectors A and B
(2) When the directions of the non-zero vectors A and B are opposite and the moduli of the vectors A and B are not equal, the direction of the vector A+vector B is the same as the direction of the vector whose moduli are large.
When the directions of the non-zero vectors A and B are opposite and the moduli of the vectors A and B are equal, the vector A+vector B is with arbitrary directions.

Wrong
(1) When the directions of the non-zero vectors A and B are the same, the direction of the vector A+vector B is the same as the direction of the vectors A and B
(2) When the directions of the non-zero vectors A and B are opposite and the moduli of the vectors A and B are not equal, the direction of the vector A+vector B is the same as the direction of the vector whose moduli are large.
When the directions of non-zero vectors A and B are opposite and the moduli of vectors A and B are equal, vector A+vector B is a zero vector with arbitrary directions.