Given plane vector A =(1,2), B=(-1, m) if A⊥ B, then the real number m equals () A.2 B.1 2 C.-1 2 D.-2 Given plane vector A =(1,2), B=(-1, m), if A⊥ B, then the real number m equals () A.2 B.1 2 C.-1 2 D.-2

Given plane vector A =(1,2), B=(-1, m) if A⊥ B, then the real number m equals () A.2 B.1 2 C.-1 2 D.-2 Given plane vector A =(1,2), B=(-1, m), if A⊥ B, then the real number m equals () A.2 B.1 2 C.-1 2 D.-2

∵

A⊥

B,

A•

B =0.
-1×1+2M=0, m=1
2.
Therefore: B.

∵

A⊥

B,

A•

B =0.
-1×1+2M=0, m=1
2.
Selected from: B.

If the module length of vector a =1, the module length of vector b =2, and the module length of vector a-b =2, what is the module length of vector a+b? (2) Given the module length of vector a =4 and the module length of vector b =3, when vector a is parallel to vector b and vector a is perpendicular to vector b, what is vector a multiplied by vector b?

1.|A|=1: a2=1|b|=2: b2=4
|A-b|=2, so there is:(a-b)2=4
I.e. a2-2ab + b2=4
Available:2ab=1
(A+b)2=a2+2ab+b2
=1+1+4
=6
|A+b|=√6
2. Available:|a|=4,|b|=3
When vector a is parallel to vector b, cos=1
So: ab=|a||b|cos=3x4x1=12
When vector a is perpendicular to vector b, there is: cos=0
So: ab=|a||b|cos=3x4x0=0

The modulus of a vector plus b vector is greater than or equal to the modulus of a vector minus b vector A greater than is always right, so what if the a vector and the b vector are perpendicular?

Convert to
(A vector + b vector)^2-(a vector - b vector)^2 Comparison:
(A vector + b vector)^2-(a vector - b vector)^2=4ab vector
4Ab vector >=0
4Ab vector =0 is perpendicular to vector a and vector b

If the module of a vector plus b vector is equal to 3 and the module of a vector minus b vector is equal to 4, what is the dot product of a vector and b vector?

Square of modulus of a+b=(a+b)2=a2+2ab+b2=9
Square of module of a-b=(a-b)2=a2-2ab+b2=16
Subtract 4ab=-7 from the above two expressions, and the point product of ab is -7/4.

Square of module of a+b=(a+b)2=a2+2ab+b2=9
Square of module of a-b=(a-b)2=a2-2ab+b2=16
Subtract 4ab=-7 from the above two expressions, and the point product of ab is -7/4.

Given that the module of vector a=2, the module of vector b=1, and the module of vector a-vector b=2, what is the vector a*vector b equal to? What is the modulus of vector a + vector b?

│A│=2
│B│=1
│A-b│=2

A*b=(a2+b2-(a-b)2)/2=(│a│2+│b│2-│a-b│2)/2=(4+1-4)/2=1/2
│A+b│=√(│a│2 b│2+2a*b)=√(4+1+1)=√6

Given points A (-1,-1), B (-4,2), C (3,0), point P is on a straight line AB and satisfies that the module of vector AP is equal to the module of vector AB by 1⁄3 times. It is known that points A (-1,-1), B (-4,2), C (3,0), and point P are on the straight line AB, and the module of vector AP is equal to the module of vector AB by 1⁄3 times, and point Q is the midpoint of line segment PC, and the coordinates of point Q are obtained. Given points A (-1,-1), B (-4,2), C (3,0), point P is on a straight line AB and satisfies that the module of vector AP is equal to the module of vector AB by 1⁄3 times. Given points A (-1,-1), B (-4,2), C (3,0), point P is on the straight line AB, and the module of the vector AP is equal to the module of the vector AB by 1⁄3 times, point Q is the midpoint of the line segment PC, and the coordinates of point Q are obtained.

A (-1,-1), B (-4,2)
Vector AB=(-3,3)
|Vector AP |=(1/3)|Vector AB |
Vector AP=(1/3) vector AB=(-1,1)
A (-1,-1)
P (-2,0)
C (3,0), point Q is the midpoint of the segment PC
Q (1/2,0)