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Given |a|=3,|b|=5, and a*b=12, how much is the projection of vector a in the direction of vector b?
A*b=|a||b|cos
The projection of vector a in the direction of vector b is |a|cos.
|A|cos=(a*b)/|b|=12/5=2.4
A*b=12, and b=5, the projection of vector a in the direction of vector b Both a and b are vectors If there is a specific process, it is best to start with the most basic
The product of two vectors is equal to the cosine of their modulus (absolute value) multiplied by their included angle, i.e. a×b=|a b|cosα(α is the included angle)
1. What is the product of two non-zero vectors a and b?
Defined as a•b = a b cos
The product of zero vector and any vector is 0, i.e.0•a=0.
2. What is the geometric meaning of the number product of plane vectors?
If the definition: b cos is called the projection of vector b in the a direction, then the geometric meaning of the vector quantity product is that the quantity product a•b is equal to the product of the length a of a and the projection b cos of b in the a direction.
A*b=12, and |b|=5, then the projection of vector a in vector b is b cosα=a*b/b=|a b|cosα/b=12/5=2.4
Known vector A, B Satisfies| B |=2, A vs. B is 60°, then B In The projection on a is ______. Known vector A, B Satisfies | B |=2, A and B is 60°, then B In The projection on a is ______. Known vector A, B Satisfies| B |=2, A and B is 60°, then B In The projection on a is ______.
0
[Urgent~] Given vector:|a|=2,|b|=3, and ab=4, then the projection of b on a is _______ I can find the cosine of the angle,
Ab=a*b*cosr=6 cosr=4.
3Sinr=3*root 5/3=root 5
Given that O, A, B are three points on the plane, there is a point C on the straight line AB, satisfying 2AC+CB=0, then OC is equal to? Given that O, A, B are three points on the plane, there is a point C on the straight line AB, satisfying 2AC+CB=0, then OC equals?
2(OC-OA)+(OB-OC)=0
2OC-2OA+OB-OC=0
0C =2OA-OB
Given that O, A, B are three points on the plane, there is a point C on the straight line AB, satisfying vector 2AC+vector CB=0, then vector OC= I' you! Given that O, A, B are three points on the plane, there is a point C on the straight line AB, satisfying vector 2AC+vector CB=0, then vector OC= I beg you!
2AC+CB=0
2(OC-OA)+OB-OC=0
2OC-2OA+OB-OC=0
OC =2OA-OB
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