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Given that the angle between vector a and b is 30°, and |a|= root number 3,|b|=1, find the cosine of the angle between two vectors a+b, a-b Given that the angle between vector a and b is 30°, and |a|= root number 3,|b|=1, find the cosine value of the angle between two vectors a+b, a-b
By the title:|a|=sqrt (3),|b|=1,=π/6, so: a·b=|a b|* cos (π/6)=sqrt (3)*sqrt (3)/2=3/2|a+b|^2=(a+b)·(a+b)=|a b||b|^2+2a·b=3+1+3=7, so:|a+b||=sqrt (7)|a-b|^2=(a-b)·(a-b)=|a b|^2-2a·b=3+1-3=1, so:|a-b|...
Given vector a=(3,2),|b|=root 2, and cosine value of angle between vector a and vector b is (root 26)/26, find vector b
Let b=(x, y)
Then: x2+ y2=2 1
A*b=|a b|*cosθ=1
That is,3x+2y=1 2
It is obtained from 1 2: x=1, y=-1 or x=-7/13, y=17/13
So vector b=(1,-1) or b=(-7/13,17/13)
Let b=(x, y)
Then: x2+ y2=2 1
A*b=|a b|*cosθ=1
That is,3x+2y=1 2
From the 1 2 solution, x=1, y=-1 or x=-7/13, y=17/13
So vector b=(1,-1) or b=(-7/13,17/13)
The CM vector =(4/3,4/3,4 times root number 2). The FD vector =(0,-2,- root number 2).
|CM vector|=8√5/3,|vector FD|=√6, vector CM*vector FD=-32/3=|vector CM vector FD|COSα, get cosα=-2√30/15
|CM vector |=8√5/3,|vector FD |=√6, vector CM*vector FD=-32/3=|vector CM |*|vector FD |α=-2√30/15
What is the cosine of the angle between vector a=(0,1,0) and b=(-3,2, root number 3)
Cosine of included angle =ab/|a||b|=2/(1*4)=1/2
Cosine of angle =ab/|a||b|=2/(1*4)=1/2
Given the plane vector a, b,|a|=1,|b|=2, and |2a+b|=Root 10, what is the angle between vector a and a-2b?
|2A+b|= square of both sides of root 10, resulting in ab=0.5
Cosc=a*(a-2b)/|a a-2b|
The numerator is zero, so it's 90 degrees.
|2A+b|= square of both sides of root sign 10, resulting in ab=0.5
Cosc=a*(a-2b)/|a a-2b|
The numerator is zero, so it's 90 degrees.
Given vector |a|=1 and vector |b|=3,|2a+b|=root number 7, find the angle between vector a and vector b
|2A+b |=√7
Square it and get
|2A+b|^2=7
4|^2+4A·b 2=7
|A|=1,|b|=3
4×1+4A·b+9=7
4A·b=-6
A·b=-3/2
Cos=(a·b)/|a b|)=(-3/2)/(1×3)=(-3/2)/3=-1/2
RELATED INFORMATIONS
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