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Given vector a=(4,2), find the coordinates of the unit vector perpendicular to a and find cosα
Let b=(x,1) be the vector perpendicular to vector a, then
A*b=4x+2=0,
X=-1/2.
|B |=√[(-1/2)^2+1]=√(5/4),
The unit vector perpendicular to vector a = soil b/|b|=(-(√5)/5,(2√5)/5), or ((√5)/5,(-2√5)/5).
Cosa=|a||b|/a•b
Given a =(4,2), the coordinate of the unit vector perpendicular to a is. Why X Y =1
The length of the unit vector is one, set the vector to (x, y)
X2+ y2=1,4x+2y = o
The vector is (√5/5,-2√5/5)
The length of the unit vector is one, set the vector to (x, y)
X2+ y2=1 and 4x+2y = o
The vector is (√5/5,-2√5/5)
Given a=(2,4), the coordinate of the unit vector perpendicular to a is The answer is as follows: To make a vector perpendicular to (4,2), the vector is collinear with (-2,4) Set to n (-2,4), where n is a constant Because is a unit vector And the mold length of (-2,4) is 2√5 So n =√5/10 So coordinates are (-√5/5,2√5/5) or (√5/5,-2√5/5 In the penultimate step, why is n equal to √5/10 and why are there so many coordinates?
This solution is not very good. Let me give you a solution. Let the vector be (m, n). Then according to the meaning:2m+4n=0m^2+n^2=1, the solution is: m=-√5/5n=2√5/5 or m=-√5/5n=-2√5/5 The coordinate of the unit vector perpendicular to a is (-√5/5,2√5/5...
This solution is not very good. Let me give you a solution. Let the vector be (m, n). According to the meaning:2m+4n=0m^2+n^2=1, the solution is: m=-√5/5n=2√5/5 or m=-√5/5n=-2√5/5 The coordinate of the unit vector perpendicular to a is (-√5/5,2√5/5...
Given vector a=(5,4) and vector b=(3,2), the unit vector in the same direction of 2a-3b is Given vector a=(5,4), vector b=(3,2), then the unit vector of 2 vector a-3 vector b in the same direction is Given vector a=(5,4) and vector b=(3,2), the unit vector in the same direction of 2a-3b is Given vector a=(5,4), vector b=(3,2), then the unit vector of 2 vectors a-3 vector b in the same direction is
2A-3b=(10,8)-(9,6)=(1,2)
|2A-3b |= Root (1+4)= Root 5
Therefore, the unit vector in the same direction is (1/root5,2/root5), i.e.(root5/5,2root5/5)
Given|vector a|=4,|vector b|=3,(2a-3b)(2a+b)=61 1. Find the angle θ between vector a and vector b 2. Find |vector a+vector b| and |vector a-vector b| Given |vector a|=4,|vector b|=3,(2a-3b)(2a+b)=61 1. Find the angle θ between vector a and vector b 2. Find |vector a+vector b| and |vector a-vector b|
(2A-3b)(2a+b)=4a^2-4ab-3b^2=64-27-4ab=61 ab=-6cosθ=ab/|a||b|=-6/12=-1/2, so θ=120 degrees|a+b|^2=a^2+2ab+b^2=16-12+9=13, so |a+b||=√13|a-b|^2=a^2-2ab+b^2=16+12+9=37|a-b||=√37
If a=(5,4) b=(3,2), then the unit vector parallel to 2a-3b is?
2A-3b=(10,8)-(9,6)=(1,2)
|2A-3b |= Root (2^2+1)= Root 5
The unit vectors parallel to 2a-3b are (1/Root 5,2/Root 5) or (-1/Root 5,-2/Root 5)
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