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Given that a, b are two vectors in the same plane, where a =(1,2),|b|= root 5/2 and a+2b is perpendicular to 2a-b, find the angle between a and b?
(A+2b)⊥(2a-b)(a+2b)●(2a-b)=0, i.e.2|a|-2|b 3a●b=0 Vectors a=(1,2)|a|=√5|b|=√5/2 2×5-5/2+3a●b=0 a●b=-5/2, i.e. cos=(a●b)/(|a||b|)=-(-5/2)/(√5×√5/2)=-1 The included angle between a and b=180° if there is any ambiguity, we will discuss it again.
The vectors A, B are known to be two vectors in the same plane, where A =(1,2), the modulus of B =(root number 5)/2, and (2A-B)⊥(A+2B) (1) Find the angle between A and B (2) Find the module of (A-B)
If the product of two vertical vectors is zero, open the product of two vertical vectors to obtain a sequence of expressions:2×IaI^2+3×IaI×IbI×The cosine value of the included angle between the two vectors is -2×IbI^2=0. It is easy to know that the module of a vector and the module of b vector can be substituted to obtain the cosine value, and then the angle can be found.
Given vector a, vector b is two vectors in the same plane, where vector a=(1,2), module of vector b=root 5/2, and a+2b is found perpendicular to 2a-b 1. A vector · b vector 2.|A-b vector|
1 A+2b is perpendicular to 2a-b (a+2b)●(2a-b)=0, i.e.2|a|2-2|b|2+3a●b=0 vector a=(1,2), module of vector b=root√5/2 2×5-5/2+3a●b=0 a●b=-5/22|a-b|2=|a|2 b|2-2a●b=5+5/4-2×(-5/2)=45/4 a-b|=3√5/...
Set vector A =(1,2), B=(2,3), if vector λ A+ B and vector C=(-4,-7) is collinear, then the value of real number λ is () A.1 B.2 C.3 D.3 2
Vector
A =(1,2),
B=(2,3), so vector λ
A+
B=(2,2 3),
Then by vector λ
A+
B and vector
C=(-4,-7) is collinear, and -4(2 3)-(2)(-7)=0,
Solution λ=2,
Therefore, B.
Vector
A =(1,2),
B=(2,3), so vector λ
A+
B =(2,2 3),
Then by vector λ
A+
B and vector
C=(-4,-7) is collinear, and -4(2 3)-(2)(-7)=0,
Solution λ=2,
Therefore, B.
Vector
A =(1,2),
B=(2,3), so vector λ
A+
B=(2,2 3),
From vector λ
A+
B and vector
C=(-4,-7) is collinear, and -4(2 3)-(2)(-7)=0,
Solution λ=2,
Therefore, B.
Let vector a=(1,2), b=(2,3) if vector λa+b is collinear with vector c=(-4,-7), then the value of λ is taken as the process second. Let vector a=(1,2), b=(2,3), if vector λa+b is collinear with vector c=(-4,-7), then the value of λ is taken as the process second.
A =(1,2), b =(2,3)
Then λa+b=(2,2 3)
Because λa+b is collinear with vector c=(-4,-7)
So (2)*(-7)-(-4)*(2 3)=0
Solution λ=2
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Set vector A =(1,2), B=(2,3), if vector λ A+ B and vector C=(-4,-7) is collinear, then the value of real number λ is () A.1 B.2 C.3 D.3 2
Vector
A =(1,2),
B=(2,3), so vector λ
A+
B=(2,2 3),
Then by vector λ
A+
B and vector
C=(-4,-7) is collinear, and -4(2 3)-(2)(-7)=0,
Solution λ=2,
Therefore, B.
Vector
A =(1,2),
B=(2,3), so vector λ
A+
B =(2,2 3),
Then by vector λ
A+
B and vector
C=(-4,-7) is collinear, and -4(2 3)-(2)(-7)=0,
Solution λ=2,
Therefore, B.
Vector
A =(1,2),
B=(2,3), so vector λ
A+
B=(2,2 3),
From vector λ
A+
B and vector
C=(-4,-7) is collinear, and -4(2 3)-(2)(-7)=0,
Solution λ=2,
Therefore, B.
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