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(4A-5b)2-2(4a-5b)(3a-2b) Calculation
Thanks.
Let vector a=(1,2), b=(m,1), if (a+2b)//(2a-b), find the product of vectors a and b RT
A+2b=(1+2m,2+2)=(1+2m,4)
2A-b=(2-m,4-1)=(2-m,3)
If (a+2b)//(2a-b),3(1+2m)-4(2-m)=0
3+6M-8+4m=0
M=1/2
A =(1,2), b =(1/2,1)
A·b=1*1/2+2*1=2.5
Given |a|=3,|b|=4, and the vector a is perpendicular to b, then the quantity product (3a-2b)*(2a+b)=?
(3A-2b)*(2a+b)=6a^2+3ab-4ab-2b^2=6|a|^2-2||b|^2=6*9-2*16=22
Given vector a=(3,-6), b=(-1,4), find |a+b|, quantity product a·b and (2a+b)·(2a-b)
Because a =(3,-6), b =(-1,4),|a|= root number (3^2+(-6)^2)=3 root number 5,|b|= root number ((-1)^2+4^2)= root number 17a·b =3*(-1)+(-6)*4=-27. Also |a+b |^2=(a+b)^2= a^2+b^2+2ab =45+17-2*27=8|a+b |=2 root number 2. And (2a+b)(2a-b)=4a^2-...
It is proved that if a vector and b vector are collinear, then 2a vector-b vector is collinear with a vector
A vector and b vector are collinear
Then there is t so that b=ta
2A-b=2a-ta=(2-t) a
So 2a-b vector is collinear with a vector
A vector and b vector are collinear
Then there is t such that b=ta
2A-b=2a-ta=(2-t) a
So 2a-b vector is collinear with a vector
Given that vectors a, b are not collinear and (a+b) are perpendicular (2a-b) and (a-2b) are perpendicular (2a+b), find the cosine of the angle between vectors a and b?
(A+b) vertical (2a-b), so a^2-2b^2+a*b=0(a-2b) vertical (2a+b), so 2a^2-2b^2-3a*b=0 can be set as a modulus 1, the former equation ×3 plus the latter equation 5a^2-8b^2=0, b^2=5/8, bring this with a^2=1 into a^2-2b^2+a*b=0, a*b=1/4, so the cosine value of included angle is a*b=1/4 divided by a, b...
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