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Vector b=(-3,1), c=(2,1), if vector a is collinear with c, find the minimum value of modulus of b+a Vector b=(-3,1), c=(2,1), if vector a is collinear with c, find the minimum value of module of b+a
A is collinear with c, so let a=k (2,1)=(2k, k)
B+a=(2k-3, k+1),|b+a|^2=(2k-3)^2+(k+1)^2=5k ^2-10k+10=5(k-1)^2+5≥5, so the minimum value of |b+a| is √5
Vector) a and b are collinear, b and c are collinear, then a and c are collinear?
Not necessarily
For example, vector a is collinear with zero vector, vector 0 is collinear with b, but a and b are not necessarily collinear
If a, b, c are all non-zero vectors, then a, c must be collinear
Let two vectors a=(2,2-cos^2α) and b=(m, m/2+sinα), where λ, m,α are real numbers. Let two vectors a=(2,2-cos^2α) and b=(m, m/2+sinα), where λ, m,α are real numbers. If a=2b, then the range of λ/m is __? "Trouble detail" Thank you so much!
2=2M
2-Cos^2α=m+2sinα
M^2-9m+4= range of cos^2α+2sinα=-(sinα-1)^2+2[-2,2]
Solve 1/4
Given a, b are two vectors, where a =(1,2) if b =(1, m) and a+2b is perpendicular to a-2b, find the value of m; find the cosine value of the included angle Q of a and b satisfying (1)
A+2b=(3,2+2m), a-2b=(-1,2-2m). Because the two vectors are perpendicular,(3/2+2m)(-1/2-2m)=-1 gives m =1/2 or -1/2 with an angle of 90 degrees or 37 degrees
A+2b =(3,2+2m), a-2b =(-1,2-2m). Because the two vectors are perpendicular,(3/2+2m)(-1/2-2m)=-1 gives m =1/2 or -1/2 with an angle of 90 degrees or 37 degrees
Given the vector AB=(3,4), the coordinate of point A is (1,2) and the coordinate of point B is obtained Given the vector AB=(3,4), the coordinates of point A are (1,2) and the coordinates of point B are obtained
AB vector = coordinates of B - coordinates of A
So B's coordinates are (4,6)
Given a=(2,-1,3), b=(-1,4,-2), c=(7,5, x), if a, b, c are coplanar, then the real number x equals
C = k1a + k2b
(7,5, X)=(2k1-k2,-k1+4k2,3k1-2k2)
=>
7=2K1-k2(1) and
5=-K1+4k2(2) and
X =3 k1-2 k2(3)
(1) + 2(2)
7 K2=17,=> k2=17/7
K1=16/7
From (3)
X =3(16/7)-2(17/7)
= 2
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