Space vector concept problem The three vectors ABC are not coplanar and form the base. But doesn't it mean that any two vectors are coplanar? Does ABC not coplanar here mean that two of the three are coplanar?

Space vector concept problem The three vectors ABC are not coplanar and form the base. But doesn't it mean that any two vectors are coplanar? Does ABC not coplanar here mean that two of the three are coplanar?

You're right. Any two vectors are coplanar,
The fact that ABC is not coplanar does not mean that two of the three are coplanar
Are three vectors that cannot be translated to the same plane

Given a (- 2,1). B (3, - 2), vector am = 1 / 5, vector AB, the coordinate of point m is

(-1,2/5)

It is proved that if M > N, A1 A2 ···· am is linearly correlated It's hard to organize language

Knowledge points:
The necessary and sufficient condition for the linear correlation of A1, A2 ···· am is that the homogeneous linear equations (A1, A2 ··· AM) x = 0 have nonzero solutions
Because R (A1, A2 ···· AM)

Given △ ABC and point m, for any point O in space, vector om = 1 / 3 (vector OA + vector ob + vector OC), if vector AB + vector AC = m, vector am Then M equals. I can do more than half of it Let the midpoint of AB be g. I will do 2 / 3 vector og plus 1 / 3 vector OC = OM / then how om shoots to the point of triangle ABC,

1. Ma, MB and MC are coplanar
Just prove that Ma + MB + MC = 0
MA=OA-OM
MB=OB-OM
MC=OC-OM
MA+MB+MC=OA+OB+OC-3OM=0
In fact, the first question can judge that ABCM is coplanar. The second question is to prove that point m should be within ABC
If point m is outside, the sum of any two vectors cannot be opposite to the direction of the third vector
So this m-point is included

If the point m is in the plane ABC and the vector om = 1 / 3 (vector OA) + 1 / 4 (vector OB) + y (vector OC) is satisfied for any point O in space, then y =?

When ABC three points are not collinear, y = 1 - 1 / 3 - 1 / 4 = 5 / 12

Let o be the outer center of triangle ABC, and point m satisfy vector OA + vector ob + vector OC = vector OM, then M is the ()? A center of gravity, B center of gravity, C center of gravity

orthocenter
AM·BC
=(OM-OA)·(OC-OB)
=(OC+OB)·(OC-OB)
=OC^2-OB^2
=|OC|^2-|OB|^2
=0
So am ⊥ BC
Similarly, BM ⊥ AC can be obtained, so that M is the vertical center