Let two vectors e1, e2 satisfy |e1|=2,|e2|=1, e1, e2 have an angle of 60 degrees, if the angle between vector 2te1+7e2 and vector e1+te2 is obtuse, Value range of factor t

Let two vectors e1, e2 satisfy |e1|=2,|e2|=1, e1, e2 have an angle of 60 degrees, if the angle between vector 2te1+7e2 and vector e1+te2 is obtuse, Value range of factor t

E1*e2=|e1 e2|* cos60°=1
The angle between vector 2te1+7e2 and vector e1+te2 is obtuse
Then (2te1+7e2)*(e1+te2)<0
2T (e1)^2+(2t^2+7) e1e2+7t (e2)^2<0
8T+2t^2+7+7t <0
2T^2+15t+7<0
-7 When vector 2te1+7e2 is collinear with vector e1+te2
Let 2te1+7e2=a (e1+te2)=ae1+ate2
2T=a,7=at
Solution t=14/2
Therefore, the angle between vector 2te1+7e2 and vector e1+te2 is the value of obtuse angle t (-7,-√14/2)∪(-√14/2,-1/2)

One math: vector e1 and e2 are known to be unit vectors with an angle of 60°, and a=2e1+e2, b=-3e1+2e2, find |a+b| and |a-b| One high mathematics: the vector e1 and e2 are known to be unit vectors with an angle of 60°, and a=2e1+e2, b=-3e1+2e2, find |a+b| and |a-b|

Make a unit circle o, take ox=1; then make the angle between oy=1 and ox as 60°; vector e1=ox=(1,0); e2=oy=(0.5,√3/2)
A=2e1+e2=2(1,0)+(0.5.√3/2)=(2.5,√3/2)
B=-3e1+2e2=(-3,0)+(5,√3)=(2,√3)
A+b=(4.5,3√3/2)|a+b|=√(4.5^2+27/4)=√27=3√3
A-b=(0.5,-√3/2)|a-b|=√(0.5^2+3/4)=1
I.e.|a+b|=3√3
|A-b |=1

Suppose that the two vectors e1e2 satisfy the absolute value of e1=2 the absolute value of e2=1 and the included angle between e1 and e2 is 60 degrees, If the included angle between 2te1+7e2 and vector e1+te2 is obtuse, calculate the value range of t Let two vector e1e2 satisfy that absolute value of e1=2 e2=1, the angle between e1 and e2 is 60 degrees, If the included angle between 2te1+7e2 and vector e1+te2 is obtuse, calculate the value range of t

The key to this problem is to calculate
It is only necessary that 2te1+7e2 is multiplied by the vector e1+te2 by less than 0, because the vector multiplication =2 vector's moduli*COS their angle. If the modulus is positive and the COS is obtuse, then the COS should be negative, then (2te1+7e2)(e1+te2)

The key to this problem is to calculate
If 2te1+7e2 is multiplied by vector e1+te2 e1+te2 by less than 0, because the vector multiplication =2 vector's modulus * COS their angle. If the modulus is positive and COS is obtuse, then COS should be negative, then (2te1+7e2)(e1+te2)

The key to this problem is to calculate
It is only necessary that 2te1+7e2 is multiplied by the vector e1+te2 by less than 0, because the vector multiplication =2 vector's moduli * COS their angle. If the modulus is positive and COS is obtuse, then COS should be negative, then (2te1+7e2)(e1+te2)

Let e1 and e2 be two unit vectors with an angle of 60 degrees. Try to find the angle of vector a=2e1+e2b=-3e1+2e2

E1e2=|e1||e2|cos60=1/2a^2=(2e1+e2)^2=4e1^2+4e1e2+e2^2=4+2+1=7, so |a||=√7b^2=(-3e1+2e2)=9e1^2-12e1e2+4e2^2=9-6+4=7, so |b||=√7ab=(2e1+e2)(-3e1+2e2)=-6e1^2+e1e2+2e2^2=-6+1/2+2=-7/2cos=ab/|a||b||=-...

Let two nonzero vectors e1 and e2 be not collinear, if vector AB=2e1+3e2, vector BC=6e1+23e2, vector CD=4e1-8e2, prove that A, B, D are collinear.

Vector BD =vector BC+vector CD=6e1+23e2+4e1-8e2=10e1+15e2=5x (2e1+3e2)=5 times of vector AB

Let e1 and e2 be two non-collinear non-zero vectors, if vector AB=e1+e2, vector BC=2e1+8e2, vector CD=3(e1-e2), prove that A, B, D are collinear Fact the value of the number k so that the vectors ke1+e2 and e1+ke2 are collinear

Prove that vector AD = vector AB + vector BC + vector CD =6(e1+ e2)=6 vector AB,
Vector AB//Vector AD, vector AB and vector AD have common point A.
A, B and D are collinear.

Prove that vector AD = vector AB + vector BC + vector CD =6(e1+ e2)=6 vector AB,
The vector AB // vector AD, vector AB and vector AD have the common point A.
A, B and D are collinear.