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The module of vector a=3, the module of vector b is equal to 4, and the positional relationship between vector a+3/4b and vector a-3/4b is The square of the module of (a+3/4b)×(a-3/4b)= a is equal to 0, so the position relation of the vector a+3/4b and a-3/4b is vertical. But the vector multiplication should not be the module of a+3/4b × the module of a-3/4b × the COS of their included angle. Besides, if the square difference is equal to zero, why are they vertical?
You can find the answer to your first question in a book, and the teacher said in class that it is a theorem and can be inferred.
The result of the square difference is 0, which shows that the module of a+3/4b × the module of a-3/4b × the COS of their included angle is 0, but the module can not be 0, which can only be the cos value of the included angle is 0, so the included angle must be 90
You can find the answer to your first question in a book, and the teacher said in class that it is a theorem and can be inferred.
The result of the square difference is 0, which shows that the modulus of a+3/4b × the modulus of a-3/4b × the COS of their included angle is 0, but the modulus can not be 0, which can only be the cos value of the included angle is 0, so the included angle must be 90
Given vector |a|=2|b|=5a·b=-3, then |a+b| equals? If |a|=2|b|=5a·b=-3, then |a+b| equals? | A| represents the modulus of the vector a | a+b| represents the absolute value vector a+vector b
|A+b|=√(a+b)2=√(a2+b2+2a·b)=√(|a|2 b|2+2a·b)=√23
As shown in the figure, M and N are the edge AD of the quadrilateral ABCD, the midpoint of the BC edge, and G is the midpoint of the MN.1. MN vector =1/2(AB vector + DC vector) As shown in the figure, M and N are respectively the edge AD of the quadrilateral ABCD, the midpoint of the BC edge, and G is the midpoint of the MN. As shown in the figure, M and N are the edge AD of the quadrilateral ABCD, the midpoint of the BC edge, and G is the midpoint of MN.1. MN vector =1/2(AB vector + DC vector)
MA vector + AB vector + BN vector = MN vector 1
MD vector + DC vector + CN vector = MN vector 2
Because MA vector + MD vector =0, BN vector + CN vector =0[ because they are the same size, opposite direction, sum to 0]
So 1+2= AB vector + DC vector =2 MN vector, so MN vector =1/2(AB vector + DC vector)
As shown in the figure, in the trapezoidal ABCD, AD//BCMN is the midpoint of waist AB, DC (1) MN//BC (2) MN=1/2(bc+ad) As shown in the figure, AD//BCoidal ABCD, AD//BCMN is the midpoint of waist AB, DC (1) MN//BC (2) MN=1/2(bc+ad)
The AN was extend to BC extension line at point E,
△ADN ECN,
AD=EC,
AN=EN,
MN is the median line of △ABE,
MN//BE, i.e. MN//BC,
MN=1⁄2BE=1⁄2(BC + CE)
=½﹙ AD+BC).
In parallelogram ABCD, MN is the midpoint of DC and BC respectively. Let AB vector equal to b vector, AD vector equal to a vector, AM vector, etc. For the m vector, the AN vector is equal to the n vector. Based on m, n, representing AB vector
Add vector symbols to all of the following:
Because AD+DM=AM, AB+BN=AN
So a+1/2b=mb+1/2a=n
B=(4n-2m)/3
Let two vectors e1, e2 satisfy |e1|=2,|e2|=1, the angle of e1, e2 is 60 degrees, if the angle of vector 2te1+7e2 and vector e1+te2 is Let two vectors e1, e2 satisfy |e1|=2,|e2|=1, the included angle of e1, e2 is 60 degrees, if the included angle of vector 2te1+7e2 and vector e1+e2 is an acute angle, the value range of factor t is obtained. Let two vectors e1, e2 satisfy |e1|=2,|e2|=1, e1, e2 have an angle of 60 degrees, if the angle between vector 2te1+7e2 and vector e1+te2 is Let two vectors e1, e2 satisfy |e1|=2,|e2|=1, the included angle of e1, e2 is 60 degrees, if the included angle of vector 2te1+7e2 and vector e1+e2 is acute, the value range of factor t is obtained. Let two vectors e1, e2 satisfy |e1|=2,|e2|=1, the angle of e1, e2 is 60 degrees, if the angle of vector 2te1+7e2 and vector e1+te2 is Let two vectors e1, e2 satisfy |e1|=2,|e2|=1, the included angle of e1, e2 is 60 degrees, if the included angle of vector 2te1+7e2 and vector e1+e2 is acute, the value range of factor t is obtained.
Let two vectors e1.e2 satisfy |e1|=2,|e2|=1, the angle between e1 and e2 is 60°, if the angle between vector 2te1+7e2 and e1+te2 is obtuse, the range of the number t is obtained.[ Solution ]|e1|=2,|e2|=1, the angle between e1 and e2 is 60°, then e12=4, e22=1, e1•e2=1, the...
RELATED INFORMATIONS
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