Given that O is the outer center of △ABC AB=2, AC=1,∠BAC=120°, let vector AB=a, vector AC=b, if vector AO=λ1a 2b, then λ1 2=

Given that O is the outer center of △ABC AB=2, AC=1,∠BAC=120°, let vector AB=a, vector AC=b, if vector AO=λ1a 2b, then λ1 2=

A (0,0), B (2,0), C (-1/2,√3/2) outer center is the intersection point of the perpendicular lines on each side of the triangle, so the center point of O (1, y) A and C is (-1/4,√3/4), and the equation of the perpendicular line is y=(√3/3)(x+1/4)3/4 The straight line crosses O, so y=(√3/3)(x+1/4)3/4=2√3/3...

Given that two non-zero vectors e1, e2 are not collinear, if ab=e1+e2, ac=2e1+e2 It is known that two non-zero vector e1, e2 are not collinear, if AB=e1+e2, AC=2e1+e2, AD=3e1-3e2, Verify a, b, c, d are coplanar (E1,e2,AB,AC,AD are all vectors) It is known that two non-zero vectors e1, e2 are not collinear if ab=e1+e2, ac=2e1+e2 It is known that two non-zero vector e1, e2 are not collinear, if AB=e1+e2, AC=2e1+e2, AD=3e1-3e2, Verify a, b, c, d are coplanar (E1,e2,AB,AC,AD are all vectors) It is known that two non-zero vectors e1, e2 are not collinear if ab=e1+e2, ac=2e1+e2 It is known that two non-zero vector e1, e2 are not collinear, if AB=e1+e2, AC=2e1+e2, AD=3e1-3e2, Verify that a, b, c, d are coplanar (E1,e2,AB,AC,AD are all vectors)

Let e1 and e2 determine the plane H, which is known from AB=e1+e2, AC=2e1+e2, AD=3e1-3e2; the plane determined by AB and AC is parallel or coincident with H. Similarly, the plane M determined by AB and AD, and the plane K determined by AC and AD are parallel or coincident with H so the four points A, B, C and D are coplanar.
Instant proof

There must be a process, every question. Let e1, e2 be non-collinear non-zero vectors and a=e1-2e2, b=e1+3e2,(1) prove that a, b can serve as a set of bases;(2) decompose the vectors c=3e1-e with a, b Another question is that if 4e1-3e2=Aa b (ab is a vector), find the value of Aμ. There must be a process, every question. Let e1, e2 be non-collinear non-zero vectors and a=e1-2e2, b=e1+3e2,(1) prove that a, b can be used as a set of bases;(2) decompose the vectors c=3e1-e with a, b Another question is that if 4e1-3e2=Aa b (ab is a vector), find the value of Aμ. There's a process. Every question. Let e1, e2 be non-collinear non-zero vectors and a=e1-2e2, b=e1+3e2,(1) prove that a, b can serve as a set of bases;(2) decompose the vectors c=3e1-e with a, b Another question is that if 4e1-3e2=Aa b (ab is a vector), find the value of Aμ.

(1) Suppose a, b can not be used as a group of bases, then a, b are collinear.
Vector a=t*vector b, i.e.
E1-2e2=te1+3te2
So t=1 and 3t=-2, which can not be true, so
A, b may serve as a set of substrates
(2) Because a=e1-2e2, b=e1+3e2
So 2a=2e1-4e2, b=e1+3e2
Add 2a+b=3e1-e2
So vector c=3e1-e2=2a+b

Suppose that two non-zero vectors e1 and e2 are not collinear, if AB=e1+e2, BC=2e1+8e2, CD=3(e1-e2) if module e1=2 module e2=3, the included angle between e1 and e2 is 60 degrees, it is determined that ke1+e2 is perpendicular to e1+ke2. Let two non non-zero vectors e1 and e2 are not collinear, if AB=e1+e2, BC=2e1+8e2, CD=3(e1-e2) if module e1=2 module e2=3, the included angle between e1 and e2 is 60 degrees, it is determined that ke1+e2 is perpendicular to e1+ke2 Suppose that two non-zero vectors e1 and e2 are not collinear, if AB=e1+e2, BC=2e1+8e2, CD=3(e1-e2) if modulo e1=2 modulo e2=3, the angle between e1 and e2 is 60 degrees, it is determined that ke1+e2 is perpendicular to e1+ke2

Because Ke1+Ke2 is perpendicular to e1+Ke2, the product between them is zero, i.e.(Ke1+Ke2)(e1+Ke2)=0, and then expanded, because e1*e1=4e1*e2=2*3*cos60=3, e2*e2=3*3*cos0=9 is substituted into the above expansion formula to obtain an equation about K, and then solved to obtain the value of K. As for AB, BC, CD is...

Let any two nonzero vectors e1, e2, e3 be non-collinear. (1) Prove that when k1e1+k2e2=0, k1=k2=0, otherwise it also holds. (2) If e1+e2 is collinear with e3 and e2+e3 is collinear with e1, find e1+e2+e3 Let any two nonzero vectors e1, e2, e3 be non-collinear. (1) It is proved that when k1e1+k2e2=0, k1=k2=0, and vice versa. (2) If e1+e2 is collinear with e3 and e2+e3 is collinear with e1, find e1+e2+e3 Let e1, e2, e3 be nonzero vectors. (1) It is proved that when k1e1+k2e2=0, k1=k2=0, and vice versa. (2) If e1+e2 is collinear with e3 and e2+e3 is collinear with e1, find e1+e2+e3

(1) Inverse proof method, if k1=0, then e1=-k2/k1*e2 is obtained from k1e1+k2e2=0, which shows that e1 and e2 are collinear and contradict with known values, so k1=0. Similarly, k2=0 can be proved. On the contrary, if k1=k2=0, there is obviously k1e1+k2e2=0.(2) Because e1+e2 is collinear with e3, there is a real number x such that...

In triangle ABC, AB=2, AC=3, D is the midpoint of BC, then vector AD is multiplied by vector BC=?

=2.5 First establish the Cartesian coordinate system with A as the origin AC set on the x-axis, then B orbit is the circle equation with A as the origin radius 2 as x2+y2=4. If B is (a, b), then B satisfies the equation. At this time, the coordinates of D are {(3+a)/2, b/2}, the vector AD is the same as the coordinates of D, and the vector BC is (3-a,-b). The vector AD multiplied by BC is (3+a)*(3-a)/...

=2.5 First establish the Cartesian coordinate system with A as the origin AC set on the x-axis, then B orbit is the circle equation with A as the origin radius 2 as x2+y2=4. Let B be (a, b), then B satisfies the equation. The coordinates of D are {(3+a)/2, b/2}, the vector AD is the same as the D coordinate, and the vector BC is (3-a,-b). The vector AD is multiplied by BC as (3+a)*(3-a)/...

=2.5 First, establish the Cartesian coordinate system with A as the origin and set AC on the x-axis, then B orbit is the circle equation with A as the origin and radius of 2 as x2+y2=4. If B is (a, b), then B satisfies the equation. At this time, the coordinate of D is {(3+a)/2, b/2}, the vector AD is the same as the coordinate of D, and the vector BC is (3-a,-b). The vector AD is multiplied by BC as (3+a)*(3-a)/...