Given the plane vector a =(1,2), vector b =(2, m), if vector a is perpendicular to vector b, then the real number m equals?

Given the plane vector a =(1,2), vector b =(2, m), if vector a is perpendicular to vector b, then the real number m equals?

Vertical is the number product is 0
I.e.2+2m=0
M=-1

Vertical is the quantity product is 0
I.e.2+2m=0
M=-1

Let vector a=(2,4), b=(1,1), if b is vertical (a+m multiplied by b), then the real number m=? Urgent Let a=(2,4), b=(1,1), if b is vertical (a+m multiplied by b), then the real number m=? Urgent

M* vector b=(m, m)
Vector a+m*vector b=(2+m,4+m)
Vector b is perpendicular to the one above, so (2+m,4+m)*(1,1)=0
So 2+m+4+m=0, so m=-3.
If you understand your question correctly, it should be.

M* vector b=(m, m)
Vector a+m* vector b=(2+m,4+m)
Vector b is perpendicular to the one above, so (2+m,4+m)*(1,1)=0
So 2+m+4+m=0, so m=-3.
If you understand your question correctly, it should be.

1. Given vector a=(2,3), vector b=(-1,2), if ma+b is perpendicular to a-2b, then real number m= 2. Given vector a modulus=3, the projection of vector b in vector a direction is 3/2, then vector a is multiplied by vector b= 3. Given vector |a|=4, vector |b|=2, the included angle between vector a and vector b is 60°, find |2a-b| 1. Given vector a=(2,3), vector b=(-1,2), if ma+b is perpendicular to a-2b, then real number m= 2. Given vector a modulus=3, the projection of vector b in vector a direction is 3/2, then vector a is multiplied by vector b= 3. Given vector |a|=4, vector |b|=2, the angle between vector a and vector b is 60°, find |2a-b| 1. Given vector a=(2,3), vector b=(-1,2), if ma+b is perpendicular to a-2b, then real number m= 2. Given vector a modulus=3, the projective of vector b in vector a direction is 3/2, then vector a is multiplied by vector b= 3. Given vector |a|=4, vector |b|=2, the angle between vector a and vector b is 60°, find |2a-b|

(1)
The vertical vector dot area is 0
Ma+b=(2m-1,3m+2)
A-2b=(4,-1)
(2M-1)*4-(3m+2)=0
5 M =6
M=6/5
(2)
A·b=|a b|*cos
=3*3/2=9/2
(3)
|2A-b|2
=4A2-4a·b+b2
=4|A|2-4a·b+|b|2
=4*42+22-4|A b|*cos60°
=64+4-4*4*2*1/2
=68-16
=52
If you still have doubts, please ask. Wish:

Given vectors a=(2,3), b=(-1,2), if λa+b is collinear with a-2b, then the real number λ is equal to?

Vector a=(2,3), b=(-1,2), if λ a+b is collinear with a-2b
Vector input a + b =(2 inputs,3 inputs)+(-1,2)=(2λ-1,3λ+2)
Vector a-2b=(2,3)-(-2,4)=(4,-1)
If λa+b is collinear with a-2b
So -1*(2in-1)=4(3in+2)
Solution: In =-(1/2)

For real numbers m and vectors a and b: m (a-b)=ma-mb;(2) for real numbers m, n and vectors a:(m-n) a=ma-na;(3) if ma=mb (m∈R), then a=b;(4) if ma=na (m, n∈R, a=0), then m=n.

3.M=0 a, b not necessarily equal

If the line x+y+m=0 intersects with the circle x^2+y^2=2 at two different points A and B, O is the coordinate origin,|vector OA+OB|>|vector OA-OB|, then the value range of the real number m Answer: If the distance from point O to straight line x+y+m=0 is greater than 1, why? If the line x+y+m=0 intersects with the circle x^2+y^2=2 at two different points A and B, O is the origin of coordinate,|vector OA+OB|>|vector OA-OB|, then the value range of real number m Answer: If the distance from point O to straight line x+y+m=0 is greater than 1, why? If the line x+y+m=0 intersects with the circle x^2+y^2=2 at two different points A and B, O is the origin of coordinates,|vector OA+OB|>|vector OA-OB|, then the value range of real number m Answer: If the distance from point O to straight line x+y+m=0 is greater than 1, why?

According to the distance d|vector OA-OB| from point O to line x+y+m=0, from the parallelogram, the diagonal of the adjacent edge with obtuse angle is shorter than the diagonal of the adjacent edge with acute angle, so the angle between vector OA and OB is acute angle.