Given 2a+b=(-4,3) a-2b=(3,4) find a*b

Given 2a+b=(-4,3) a-2b=(3,4) find a*b

2A+b=(-4,3)1
A-2b =(3,4)2
1×2+2:5A=(5,10)
So a =(1,2)
1-2×2:5B=(-10,-5)
Therefore, b=(-2,-1)
Then: a*b=-2-2=-4

Given plane vector A =(1,1), B=(1,-1), then vector 1 2 A-3 2 B=______. Given plane vector A =(1,1), B=(1,-1), then vector 1 2 A-3 2 B =______.

Plane vector

A =(1,1),

B=(1,-1),
Vector1
2

A-3
2

B=(1
2,1
2)-(3
2,−3
2)
=(-1,2).
Therefore, the answer is:(-1,2).

Plane vector

A =(1,1),

B=(1,-1),
Vector 1
2

A-3
2

B=(1
2,1
2)-(3
2,−3
2)
=(-1,2).
Therefore, the answer is:(-1,2).

Given vector a=(1,3), vector b=(4,-2), find (2a+b)·(a-2b)

2A+b=(6,4);
A-2b=(-7,7)
(2A+b)(a-2b)=-42+28=-14;
I'm happy to answer your questions,skyhunter002 answer them
If you have any questions you can ask,

2A+b=(6,4);
A-2b=(-7,7)
(2A+b)(a-2b)=-42+28=-14;
I'm happy to answer your questions,skyhunter002 answer them
If there's anything you do n' t understand,

What is the parallel formula of two vectors in space? N1=(x1, y1, z1) n2=(x2, y2, z2) The formula that proves that two vectors are parallel is?

(1) The vector parallel angle is 0 or 180, prove that
Cos (n1, n2)= n1*n2/|n1|/|n2|=1
|N1|,|n2|are modulo n1, n2, i.e. length
Or
(2) Prove that n2= c*n2, c is a real number
That is, x1=c*x2, y1=c*y2, z1=c*z2

(1) The vector parallel angle is 0 or 180, prove that
Cos (n1, n2)= n1*n2/|n1|/|n2|=1
|N1|,|n2|are modulo n1, n2, i.e. length
Or
(2) It is sufficient to prove that n2= c*n2, and c is a real number.
That is, x1=c*x2, y1=c*y2, z1=c*z2

What is the formula for judging that two vectors are parallel?

Vector a//vector b
A (x1, y1), b (x2, y2)
Vector a//vector b
Then y1/x1=y2/x2

Space vector vertical formula Urgent

A =(ax, ay, az)
B=(bx, by, bz)
A =0
B=0
If a, b is vertical, then:
1: Ab=ax×bx+ay×by+az×bz=0;
Or ab =| a || b | cos (π/2)=0;
2: The zero vector is orthogonal to any vector.