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Known vector A =(1,2), B =(x,1), and A+2 B and 2 A- B parallel, then x equals () A.1 B.-2 C.1 3 D.1 2
From the meaning of the title
A+2
B=(1+2x,4),2
A-
B=(2-x,3),
Because
A+2
B and 2
A-
B parallel, so 3(1+2x)-4(2-x)=0,
Solve the equation to get x=1
2
Therefore, D
From the meaning
A+2
B=(1+2x,4),2
A-
B=(2-x,3),
Because
A+2
B and 2
A-
B parallel, so 3(1+2x)-4(2-x)=0,
Solve the equation to get x=1
2
Therefore, D
Vector a=(1.2), b=(x.1)(1) When a+2b is parallel to 2a-b, find x (2) When a+2b is perpendicular to 2a-b, find x
1.A+2b=(1+2x,4)2a-b=(2-x,3)
3(1+2X)-4(2-x)=0x =1/2
2.(1+2X)(2-x)+12=0
X=-2 or x=7/2
Plane vector. If a=(-1,1), b=(2, m), and 2a+2b is parallel to a-2b, then the value of m is?
The answers of the first three friends, to varying degrees, have some problems. Correct as follows:
A=(-1,1), b=(2, m)
2A+2b =(2,2m+2)
A-2b =(-5,1-2m)
(2A+2b)∥(a-2b)
2(1-2M)-(-5)(2m+2)=0
2-4 M +10 m +10=0
I.e.6 m +12=0
M =-2
Operation of vector: if m (3a-2b)+n (4a+b)=2a+5b, find the value of n, n
This is called a discussion question:
1) Vector a = b =0 then n, m ∈ R
2) When a=0 b is not equal to 0
-2Mb+nb=5b-2m+n=5, m=∈R, n=(5+2m)
3) A
Not equal to 0
3Ma+4na=2a;3m+4n=2; m∈R, n=(2-3m)/4
4) A, b are not 0;
The vector can be treated as a common variable;
Solve identity m (3a-2b)+n (4a+b)≡2a+5b
Get:
(3M+4n) a=2a
(-2M+n) b=5b
That is:
3M +4n =2
-2M+n=5
Solve the equation group to obtain:
M=-18/11;
N=19/11;
If sin^4a/sin^2b+cos^4a/cos^2b=1, prove that sin^4b/sin^2a+cos^4b/cos^2a=1. )
CERTIFICATE:
Input too much trouble, use the change element method
Let A = sin2A, B = sin2B
Sin^4a/sin^2b + cos^4a/cos^2b=1
I.e. A2/B+(1-A)2/(1-B)=1
A2(1-B)+(1-A)2B = B (1-B)
A2-A2B+B-2AB+A2B=B-B2
A2-2AB=-B2
A2-2AB+B2=0
∴(A-B)2=0
A=B
Sin^4b/sin^2a+cos^4b/cos^2a
= A2/B +(1-B)2/(1-A)
= A2/A +(1-A)2/(1-A)
=A+1-A
=1
The equation holds.
If m (3a-2b)+n (4a+b)=2a-5b, find the value of m and n
3M+4n=2-2m+n=-5m=2n=-1
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