Point A (-3.2), point B (3.6), point C satisfies vector AC = vector CB, then vector AB is multiplied by vector AC =?

Point A (-3.2), point B (3.6), point C satisfies vector AC = vector CB, then vector AB is multiplied by vector AC =?

Let C (x, y) have vector AC (x+3, y+2) vector AB (6,8) vector CB (3-x,6-y)
According to the vector AC=CB, x+3=3-x; y+2=6-y.x=o, y=2
Then vector AC is (3,4) vector AC*vector A B=3*6+4*8=50

Given vector AB=(2,-1), vector AC=(-4,1), then what is vector BC equal to

(-6,2)

The AC vector is known as the sum vector of the AB vector and the AD vector, and the AC vector is equal to the a vector and the BD vector is equal to the b vector, respectively Representation of AB vector BD vector with a vector b vector

Use a vector and b vector to represent AB vector and BD vector.
Represent AB vector AD vector with a vector b vector
Solve AC=AB+AD, i.e. a=AB+AD, BD=b=AD-AB.
AD=(a+b)/2, AB=(a-b)/2.

Use a vector and b vector to represent AB vector and BD vector.
Representation of AB vector AD vector with a vector b vector
Solve AC=AB+AD, i.e. a=AB+AD, BD=b=AD-AB.
AD=(a+b)/2, AB=(a-b)/2.

The vector AB+vector AD=vector AC and vector AC=a vector BD=b are known and the vector AB vector AD is represented by a b 'Thank you'

Because AB + AD = AC
I.e. AB+AB+BD=AC
AC=a, BD=b
So AB=(a-b)/2, AD=AC-AB=(a+b)/2.

Given the AB vector = aAC vector = bBD vector =3DC vector, a, b denotes the AD vector and the AD vector is =?

0

In triangle ABC, AD is perpendicular to AB, vector BC is equal to 3 times vector BD, module of vector AD is equal to 1, and vector AC*AD In triangle ABC, AD is perpendicular to AB, vector BC is equal to 3 times vector BD, module of vector AD is equal to 1, and vector AC*AD is obtained In triangle ABC, AD is perpendicular to AB, vector BC is equal to 3 times vector BD, the modulus of vector AD is equal to 1, and vector AC*AD is obtained

AC·AD
=(AB+BC) AD
= AB·AD+BC·AD
=BC·AD
=3·(BD·AD)
=3(BDcos∠ADB·AD)
=3AD^2
=3