In a parallelogram, what process is the vector AB + vector CB-vector DC equal to

In a parallelogram, what process is the vector AB + vector CB-vector DC equal to

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It is known that A, B, and C are ord three points on that straight line L, Which vectors are in the same direction Given that A, B, and C are three points in order on the straight line L, it is indicated that in the vector AB, the vector AC, the vector BA, and the vector CB, Which vectors are in the same direction

AB and AC are in the same direction, BA and CB are in the same direction.

Given that A, B, C are three points in order on the straight line L, point out which of the vectors AB, AC, BA, CB are the vectors with the same direction Super Simple Vector Problem It is known that A, B and C are in order on the straight line L, point out which of the vectors AB, AC, BA, CB are the with the same direction Super Simple Vector Problem

ABAC is a group
BACB is a group

In triangle ABC, if D is a point on the edge of AB, if vector AD=2DB, vector CD=1/3CA CB, then λ is equal to () A.2/3 B.1/3 C.-1/3 D.-2/3 In the triangle ABC, if D is a point on the edge of AB, if vector AD=2DB, vector CD=1/3CA CB, then λ is equal to () A.2/3 B.1/3 C.-1/3 D.-2/3

λ=2/3
AD=2DB, so D is AB triple point.
Let CE=1/3CA, E be on CA, then, E is CA tri-point.
DE//CB
By the law of addition of vectors,
With CF=2/3CB such that CEDF is a parallelogram,
So λ=2/3

λ=2/3
AD=2DB, so D is AB triple point.
Let CE=1/3CA, E be on CA, then, E is CA tri-point.
DE//CB
From the law of addition of vectors,
With CF=2/3CB such that CEDF is a parallelogram,
So λ=2/3

λ=2/3
AD=2DB, so D is the triple point of AB.
Let CE=1/3CA, E be on CA, then, E is CA tri-point.
DE//CB
By the law of addition of vectors,
With CF=2/3CB such that CEDF is a parallelogram,
So λ=2/3

In triangle abc, the height of ab side is cd, vector CB = a vector, vector CA = b vector, a vector * b vector =0, and modulus of a =1, modulus of b =2, then Vector ad is represented by vector a and vector b In triangular abc, the height of ab side is cd, vector CB = a vector, vector CA = b vector, a vector * b vector =0, and modulus of a =1, modulus of b =2, then Vector ad is represented by vector a and vector b

A vector*b vector=0, CA⊥CB
By Pythagorean Theorem, AB =√5
A=∠A, two right angles are equal, ADC ACB
AD/AC=AC/AB
AD=AC2/AB=4/√5=(4√5)/5
Vector b* Vector A B=2 5*sin⊙
Vector AB = Vector CB - Vector CA = Vector a - Vector b
Vector b*(vector a-vector b)=2 5*sin⊙
Sin⊙=vector b*(vector a-vector b)/(2 5)
Vector b*Vector ad=[2*(4√5)/5]*sin⊙
Sin⊙ substituted into the formula
Vector b*Vector ad=[2*(4√5)/5]*[ Vector b*(Vector a-Vector b)/(2 5)]
Reduced to vector ad=4/5(vector a-vector b)

Vector AB=(1,3), CB=(2,2), then AC=

AC=AB-CB=(1-2.3-2)=(-1.1)