Given a=(2m-1,2+m). If the module of vector a = root number 10 and is in the same direction as b (-6,2), then the value of real number m is [] Given a=(2m-1,2+m). If the module of vector a = root number 10 and is in the same direction as b (-6,2), then the value of real number m is [] A and b are vectors (A)1(B)-1 (C)±1(D) without solution Given a =(2m-1,2+m). If the module of vector a = root number 10 and is in the same direction as b (-6,2), then the value of the real number m is [] Given a =(2m-1,2+m). If the module of vector a = root number 10 and is in the same direction as b (-6,2), then the value of the real number m is [] A and b are vectors (A)1(B)-1 (C)±1(D) without solution

Given a=(2m-1,2+m). If the module of vector a = root number 10 and is in the same direction as b (-6,2), then the value of real number m is [] Given a=(2m-1,2+m). If the module of vector a = root number 10 and is in the same direction as b (-6,2), then the value of real number m is [] A and b are vectors (A)1(B)-1 (C)±1(D) without solution Given a =(2m-1,2+m). If the module of vector a = root number 10 and is in the same direction as b (-6,2), then the value of the real number m is [] Given a =(2m-1,2+m). If the module of vector a = root number 10 and is in the same direction as b (-6,2), then the value of the real number m is [] A and b are vectors (A)1(B)-1 (C)±1(D) without solution

Because the modulus of a = sqrt ((2m-1)^2+(2+m)^2)= sqrt (5m^2+5)= sqrt (10)
Get 5m^2+5=10
Get m = plus or minus 1
But since it is in the same direction as b (-6,2)
So m=-1
Select B

Because the modulus of a = sqrt ((2m-1)^2+(2+m)^2)= sqrt (5m^2+5)= sqrt (10)
Get 5m^2+5=10
Get m = plus or minus 1
But since it is in the same direction as b (-6,2)
So m=-1
Option B

Given plane vector A =(2m+1,3) B,=(2, m), and A∥ B, the value of the real number m is equal to () A.2 or -3 2 B.3 2 C.-2 or 3 2 D.-2 7

∵

A∥

B, m (2m+1)-6=0,
2M2+m-6=0,
Solution m=3
2 Or -2.
C.

Given |a|=4,|b|=6 and the angle of ab is 60 find the angle of vector ab a (a+b) Given |a|=4,|b|=6 and the angle of ab is 60 find the angle of vector ab a (a+b)

According to the title,|a|=4,|b|=6,=π/3, so: a·b=|a||b|* cos (π/3)=4*6/2=12 According to your writing, a·b is a scalar, a·(a+b) is also a scalar, there is no included angle, unless: a (a+b)=|^2+a·b=16+12=28, and:|a+b|^2=(a+b)·(a+b)=|a...

Given vector m={ cosα-(√2)/3,-1}, vector n=(sinα,1), vector m and vector n are collinear and [-π/2,0]. (1) Find the value of sin cosα (2) Find the value of (sin2α)/(sinα-cosα)

(1) Because vector m is collinear with vector n,-sinα= cosa -(√2)/3, i.e. sinα+ cosα=(√2)/3
(2) Sin2α=(sinα+cosα)^2-1=2/9, so (sinα-cosα)^2=1-sin2α=7/9, so the original formula =2/9*9/7=2/7

Given vector m=(cosθ, sinθ) and n=(√2-sinθ, cosθ),θ belongs to (180,360), and |m+n|=8/5 2, find the value of cos (θ/2+180/8)

You don't have a clear answer to your question. I' ll probably say that according to |m+n|=8/5 2 and |m+n|=root sign (m+n)^2=root sign (m^2+n^2+2mn), the value you have in the expanded graph becomes a real number, such as sinθ^2+cosθ^2=1, and then you can combine them to get a relationship about θ(possibly a multiple angle, or...

Given vector m=(root 3sinx/4,1), vector n=(cosx/4, cos^2x/4) 1. Multiply vector m by vector n=1, find the value of cos (∏/3+x) 2. Let f (x)=vector m multiplied by vector n. In triangle ABC, the opposite sides of angles A, B, C are a, b, c, and satisfy (2a-c) cosB=bcosC.

1.M·n=√3s in (x/4) cos (x/4)+cos2(x/4)=(√3/2) sin (x/2)+(1/2) cos (x/2)+1/2=cos (x/2-π/3)+1/2=1cos (x/2-π/3)=-1/2.x/2-π/3=±2π/3+2kπ, x/2=±2π/3+2kπ+π/3x=±4π/3+4kπ+...