Given triangle ABC, point D is on the edge of BC, and vector CD = 2dB, vector CD = Rab + sac, then R + s

CD=2/3CB=2/3(AB-AC)=2/3AB-2/3AC
∴r=2/3,s=-2/3
∴r+s=0

In triangle ABC, D is on the extension line of line BC, and BC vector = 3CD vector, point O is on line CD, if Ao vector=
In △ ABC, point D is on the extension line of line BC, and BC vector = 3CD vector, and point O is on line CD (not coincident with points c and D). If Ao vector = XAB vector + (1-x) AC vector, then the value range of X is________

0 is greater than x, X is less than one third
Cause, vector co = vector Ao - vector AC
If Ao vector = XAB vector + (1-x) AC is brought in, XAB vector - x vector AC = vector Co,
Vector BC with vector co greater than 0 and less than one third
So 0 < - x (vector AC - vector AB) < 1 / 3 vector BC
The solution is 0 ＞ x ＞ - 1 / 3

If n n-dimensional vectors are linearly independent, then the determinant is not equal to 0. Why?

N n-dimensional vectors are linearly independent, which means that the rank of n-dimensional vectors is n (n maximal linearly independent group)
Since full rank, it means that the corresponding determinant is 0!

Why is the number of vectors equal to the dimension and determinant equal to 0 linearly related

Vector group A1,..., as correlation
The homogeneous linear equations x 1 a 1 +... + X SAS = 0 have nonzero solutions
When the number of vectors is equal in dimension
The homogeneous linear equations X1A1 +... + xsas = 0 have nonzero solutions
Coefficient determinants | A1,..., as | = 0 (otherwise, it is known by crammer theorem that there is a unique solution, that is, there is only a zero solution)
So the conclusion is valid

Given triangle ABC, point D is on the edge of BC, and vector CD = 2dB, vector CD = Rab + sac, then R + s