It is known that E1 and E2 are two unit vectors with an angle of 2 π / 3, a = e1-2e2, B = Ke1 + E2. If a * b = 0, what is the value of real number k A * B is the scalar product of vectors

It is known that E1 and E2 are two unit vectors with an angle of 2 π / 3, a = e1-2e2, B = Ke1 + E2. If a * b = 0, what is the value of real number k A * B is the scalar product of vectors

a=e1-2e2
,b=ke1+e2
a*b
=(e1-2e2)(ke1+e2)
=k(e1)²+(1-2k)e1e2-2(e2)²
=k+(1-2k)*(-1/2)-2
=k-(1/2)(1-2k)-2
=k+k-5/2=0
So 2K = 5 / 2
therefore
k=5/4

Let vectors E1 and E2 not be collinear, a = E1-E2, B = - E1 + E2, C = 2E1 + E2, and use vectors a and B as bases to represent vector C
Sorry, I missed a 2 in the title
Let the vectors E1 and E2 not be collinear, a = E1-E2, B = - E1 + 2e2, C = 2E1 + E2. Use vectors a and B as bases to represent vector C
sorry

Let C = XA + Yb, then:
2e1＋e2=x(e1－e2)＋y(－e1＋2e2)
2e1＋e2=(x－y)e1＋(2y－x)e2
The results are as follows
X-Y = 2 and 2y-x = 1
Then: x = 5, y = 3
The results show that C = 5A + 3B

Let vectors E1 and E2 not be collinear, a = 2e1-e2, B = - E1 + E2, C = E1-E2, and use vectors a and B as bases to represent vector C

Let C = m vector a + n vector B
∵ vector a = 2e1-e2, B = - E1 + E2, C = E1-E2
The vector C = m (2e1-e2) + n (- E1 + E2) = (2m-n) e1 + (n-m) E2
The vectors E1 and E2 are not collinear
∴2m-n=1①,n-m=-1②
The simultaneous solution of 1 and 2 gives m = 0, n = - 1
A: vector C = - B

In the triangle ABC, D is on the edge of BC, and the vector CD = - 2 vector BD. if the vector CD = P vector AB + Q vector AC., then p + Q=

From CD = - 2bd, the distance between D and B on CD is 1 / 3CB
Then CD = 2 / 3CB
CB＝AC－AB.
So CD = - (2 / 3) AB + (2 / 3) AC
p+q=0