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If the vectors a and b are not collinear, a*b is not equal to 0. And c=a-[(a*a)/(a*b)] b, the angle between a and c is? Vector,* Multiplied by point Please don't give me answers that have been answered. If the vectors a and b are not collinear, a*b is not equal to 0, and c=a-[(a*a)/(a*b)] b, the angle between a and c is? Vector,* is a point multiplication Please don't give me answers that have been answered. If the vectors a and b are not collinear, a*b is not equal to 0. And c=a-[(a*a)/(a*b)] b, the angle between a and c is? Vectors,* for dot multiplication Please don't give me answers that have been answered.
A*c=a*[ a-[(a*a)/(a*b)] b ]=a*a-[(a*a)/(a*b)](a*b)=0.The angle between a and c is 90°
A*C=A*[ A-[(A*A)/(A*B)] B ]=A*A*-[(A*A)/(A*B)](A*B)=0. The angle between A and C is 90°
If vector a and vector b are not collinear, a*b is not equal to 0, and c=a*[(a*b)/(a*a)]-b, then the angle between a and c is ___.
Cos=ac/|a||c|
Because ac=a {a*[(a*b)/(a*a)]-b}
=Aa [(ab/aa)]-ab
=0
So cos=ac/|a||c|=0
So=90
Given that vectors a and c are not collinear, vector b is not equal to 0, and (ab) c=(bc) a, d=a+c, then <向量b,向量d>=___________?
The two vector products ab and bc are quantities, and a and c are not collinear, and (ab) c=(bc) a, only ab and bc are 0, then bd=(a+c) b=ac+bc=0+0=0, the two are vertical, and the included angle is 90 degrees.
The two vector products ab, bc are quantity, and a, c are not collinear, and (ab) c=(bc) a, only ab, bc are 0, then bd=(a+c) b=ac+bc=0+0=0, the two are vertical, the included angle is 90 degrees
If the vector a and b are not collinear, a*b is not equal to zero, and c=a-(a*a) b/a*b, then what is the included angle of ac?
Let a, b be θ, vector a and b are not collinear, a=0, b=0,
C=a-[(a*a)/(a*b)] b=a-[|^2/(|a||b|cosθ)] b,1
If c =0
A|b|cosθ=|a|b
Formwork taking on both sides
Cos θ=±1
A, b are parallel. c=0
Ac=|^2-[|a|^2/(|a b|cosθ)] a·b
=|^2-[|A|^2/(|a b|cosθ)]|b a|cosθ
=0
Ac=|c||a|cosβ,β is the angle between c and a. a=0, c=0.
Cos β=0
The angle is 90°.
If vectors a, b satisfy |a|=|b|=1, the angle between a and b is 60°, then aa+ab =() If vectors a, b satisfy |a|=|b|=1, the angle between a and b is 60°, then aa+ab =() A,1/2 B,3/2 C,1+(3/2) D,2 If the vectors a, b satisfy |a|=|b|=1 and the angle between a and b is 60°, then aa+ab =() If the vectors a, b satisfy |a|=|b|=1 and the angle between a and b is 60°, then aa+ab =() A,1/2 B,3/2 C,1+(3/2) D,2
Aa+ab
= A 2 a b|*cos60°
=1+1/2
=3/2
Given vector a=(-√3,1).b=(1,0). Find the angle between ab and a, b
A·b=-√3·1+1.0=-√3
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