Given vector a, b satisfies |a|=|b|=1, real number m, n satisfies m^2+n^2=1, then the value range of |ma+nb| is |Ma+nb ma nb 2(m^2+n^2)

Given vector a, b satisfies |a|=|b|=1, real number m, n satisfies m^2+n^2=1, then the value range of |ma+nb| is |Ma+nb ma nb 2(m^2+n^2)

|Ma+nb|

Given vector a, b is not collinear, and ma+nb=0, then m^2+n^2=

Given vectors a, b are not collinear and ma+nb=0 yields m=n=0
M^2+n^2=0

Given vectors a, b are not collinear, and ma+nb=0 yields m=n=0
M^2+n^2=0

Given vector a=(1,1) b=(1,-1) c=(√2 cosa,√2 sina)(a∈R) real number m, n satisfies ma+nb=c, then (m-3)2+n2

I'm doing it.

Known vector A =(1,1), B =(1,−1), C =( 2 Cosα, 2Sinα), real number m, n satisfies m A+n B= C, the maximum value of (m-3)2+n2 is () A.2 B.4 C.8 D.16

M

A+n

B=

C,
(M+n, m-n)=(
2 Cosα,
2Sinα)( R)
M + n=
2 Cosα, m-n=
2Sinα,
M = sin (
4), N = cos (
4),
(M-3)2+n2=m2+n2-6m+9=10-6sin ()
4)
Sin (
4)∈[-1,1]
The maximum value of (m-3)2+n2 is 16
Therefore, D

Given vector a=(x1, y1) vector b=(x2, y2) prove that there is a unique real pair (m, n) such that c=ma+nb Abc is a vector. Both a and b are non-zero vectors

If k1a+k2b=0, then there must be k1=k2=0, and there must be such a condition. Let us prove the following: Suppose that there is another real number pair (m1, n1) which can also make c=m1a+n1b.

Let two vectors a=(2,2-cos^2α) and b=(m, m/2+sinα), where λ, m, a are real numbers Let two vectors a=(2,2-cos^2α) and b=(m, m/2+sinα), where λ, m, a are real numbers.

Because a=2b,2=2m,2-(cosa)^2=m+2sina,
2=2M, then λ/m=2-2/m.
Substitute λ=2m-2 into 2-cos^2α=m+2s inα to obtain:
4M^2-9m+4= range of cos^2α+2sinα=-(sinα-1)^2+2[-2,2]
I.e.-2≤4m^2-9m+4≤2,
Solution:1/4≤m≤2
Then 1/2≤1/m≤4-1≥-2/m≥-8
Thus a range of λ/m=2-2/m [-6,1]