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Given vector 2a+b=(0,-5,10), c=(1,-2,-2), a*c=4,|b|=12, then= The answer is 120 degrees.
(2A+b)*c =(0,-5,10)*(1,-2,2-2)
2A*c+b*c=-10
B*c =-18
| C |=(1^2+(-2)^2+(-2)^2)^0.5=3
Cos = b*c/(|b c|)=-18/(12*3)=-1/2;
= 120°
Given vector a =(-3,4). Vector b is opposite vector a, and vector |b|=10, then vector b =
Vector a =(-3,4). Vector b is inverse to vector a
Then vector b=-xa=-x (-3,4)=(3x,-4x)
Lbl=√9x2+16x2=5x=10
So x =2
So vector b=(6,-8)
Vector a =(-3,4). Vector b is opposite vector a
Then vector b=-xa=-x (-3,4)=(3x,-4x)
Lbl=√9x2+16x2=5x=10
So x =2
So vector b=(6,-8)
Given that a, b is a nonzero vector and satisfies (a-2b) perpendicular a,(b-2a) perpendicular b, what is the angle between a and b? Process of finding Given that a, b is a nonzero vector and satisfies (a-2b) perpendicular a,(b-2a) perpendicular b, what is the angle between a and b? Process of seeking
A (a-2b)=0=a2-2ab=(b-2a) b=b2-2ab, a2=b2.|a|=|b|
|A|2-2|a|2cos
Given the vector a=(1,1), b=(2,1), if (a b)×a=0, then the real number λ is equal to?
First of all, the title should be changed to: Given vector a=(1,1), b=(2,1), if (a b).a=0, then the real number λ is equal to?
The reason is that "X" operation is cross multiplication, and it has its own set of operation rules. Remember that point multiplication is different from cross multiplication, and it must be signed operation. Do not confuse
Because a +λ b =(1,1)+λ(2,1)=(1+2λ,1+λ)
And a b).a=(1+2λ)*1+(1)*1=3 2=0(this is the coordinate operation of the quantity product)
Finally,λ=-2/3
Given A (6,8).B (2,-3), then the coordinates of vector AB are?
2-6=-4,-3-8=-11, So (-4,-11) end point minus start point
Given A (1.-2) B (2,4) C (4.-3) D (X,1) What is the modulus of vector BD if vector AB is collinear with vector CD Given A (1.-2) B (2,4) C (4.-3) D (X,1) If the vector AB is collinear with the vector CD, what is the modulus of the vector BD
AB ={1,6}
CD={ X-4,4}
If vector AB is collinear with vector CD, then there is
1: X-4=6:4
So 6X-24=4
So X =28/6=14/3
BD={ X-2,1-4}={8/3,-3}
Then |BD|=under root (64/9+9)=under root (145/9)=[ under root (145)]/3
AB ={1,6}
CD={ X-4,4}
Vector AB is collinear with vector CD, then there is
1: X-4=6:4
So 6X-24=4
So X =28/6=14/3
BD={ X-2,1-4}={8/3,-3}
Then |BD|=under root (64/9+9)=under root (145/9)=[ under root (145)]/3
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