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Given that the module of a is 8, and e is the unit vector, the projection of a in the direction of e when the angle between them is 6 Given that the module of a is 8, and e is the unit vector, the projection of a in the direction of e when the angle between them is a p/6
A Projection in direction e
=A*e=|a|cos (π/6)=4√3.
Given vector a vector b, satisfying the absolute value of vector b=2 and the angle between vector a and vector b is 60 degrees, what is the projection of vector b on vector a? Given vector a vector b, satisfying the absolute value of vector b =2 and the angle between vector a and vector b is 60 degrees, what is the projection of vector b on vector a? Given vector a vector b, satisfying the absolute value of vector b=2 and the included angle between vector a and vector b is 60 degrees, what is the projection of vector b on vector a?
It is not the absolute value of vector b =2, but the module of vector b =2. These two concepts are different.
The projection of vector b on vector a = the modulus of vector b * cos =2* cos 60°=2*1/2=1
Under correction, not the absolute value of vector b =2, but the module of vector b =2. These two concepts are different.
The projection of vector b on vector a = the modulus of vector b * cos =2* cos 60°=2*1/2=1
Given that the vectors a, b are unit vectors with an angle of 60 degrees, the absolute value of vector a+3b is? Detailed steps! Thanks! Given that vector A and vector B are all unit vectors, their included angle is 60 degrees, then the absolute value of vector A+3B is equal to? Detailed steps! Thanks!
Come on, it's a model, not an absolute value.
(A+3b)^2= a^2+6a·b+9b^2
=1+6*1*1*Cos (60°)+9*1
=13
|A+3b|=sqrt [(a+3b)^2]=sqrt (13)
Come on, it's the mold, not the absolute.
(A+3b)^2= a^2+6a·b+9b^2
=1+6*1*1*Cos (60°)+9*1
=13
|A+3b|=sqrt [(a+3b)^2]=sqrt (13)
Known| A|=8, E is the unit vector when the angle between them is π 3 O'clock, A In The projection in direction e is () A.4 3 B.4 C.4 2 D.8+2 3 Known| A|=8, E is the unit vector when the angle between them is π At 3 o'clock, A In The projection in direction e is () A.4 3 B.4 C.4 2 D.8+2 3
From the geometric meaning of the product of two vectors:
A In
The projection in direction e is:
A•
E =|
A||
E|cosπ
3=8×1×1
2=4
Therefore, select B
From the geometric meaning of the product of two vectors:
A In
The projection in direction e is:
A•
E =|
A||
E|cosπ
3=8×1×1
2=4
Reason selection B
Given vector absolute value a=4, vector absolute value b=3, the angle between a and b is 150 degrees, then the projection of vector b on vector a is
Vector absolute value a=4, vector absolute value b=3, the angle between a and b is 150 degrees,
A*b=|a b|*cos150°=12*(-√3/2)=-6√3
The projection of vector b on vector a is =|b|*cos150°=3*(-√3/2)=-3√3/2
If the angle between the vector of four and the unit vector is two-thirds of the pie@, then the projection of the vector shoulder a in the shoulder e direction is _ If the angle between the vector of length four and the unit vector is two-thirds of@, then the projection of the vector shoulder a in the shoulder e direction is _
Is vector e a unit vector? Projection? I thought it was a projection.
The projection of vector a in vector e is |a cos (2π/3)=4×(-1/2)=-2
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