Given that a and b are both vectors _a_=10, the number of projectiles of b in the a direction is -3, multiply vector a by vector b It is bet to have a picture, Given that a and b are all vectors _a_=10, the number of projectiles of b in the a direction is -3, multiply vector a by vector b It is bet to have a picture, Given that a, b are all vectors _a_=10, the number of projectiles of b in the a direction is -3, multiply vector a by vector b It is bet to have a picture,

Given that a and b are both vectors _a_=10, the number of projectiles of b in the a direction is -3, multiply vector a by vector b It is bet to have a picture, Given that a and b are all vectors _a_=10, the number of projectiles of b in the a direction is -3, multiply vector a by vector b It is bet to have a picture, Given that a, b are all vectors _a_=10, the number of projectiles of b in the a direction is -3, multiply vector a by vector b It is bet to have a picture,

Let that angle of vector a, b be theta,
According to the definition of projective, the projective number of b in a direction is |b|cosθ.
By the known |b|cosθ=-3.
Define a•b=|a||b|cosθ=10*(-3)=-30.

Let that angle of vector a, b be theta,
According to the definition of projective, the projective number of b in a direction is |b|cosθ.
From the known |b|cosθ=-3.
Define a•b=|a||b|cosθ=10*(-3)=-30.

If the vector a=(m,1) and b=(4, m) are collinear and in the same direction, then m=

Because vector a is collinear with vector b and has the same direction
So let a=xb (and x is a positive number)(a, b represent a vector and b vector, respectively)
So m=4x,1=mx
Solution x=1/2 or x=-1/2(rounded)
So m=2

Given vectors a=(2,3), b=(-1,2), if (ma+nb) is parallel to (a-2b), then m/n equals (), A,-2 B,2 C,-(1/2) D,1/2

Vector a=(2,3), b=(-1,2),
Then (ma+nb)=(2m-n,3m+2n)
(A-2b)=(4,-1)
∵(Ma+nb) is parallel to (a-2b),
∴(2M-n)/(3m+2n)=4/(-1)
N-2m=12m+8n
14M=-7n
M/n=-1/2
Select C

Is vector a multiplied by vector b equal to 0, or vector 0?

Vector A multiplied by vector B=0
There are two possibilities:
(1) There is zero vector in A and B, that is, A=0 vector or B=0 vector.
(2) There is no zero vector in A and B. When vector A and B are perpendicular, vector A multiplies vector B=0.
That is, when vector A is multiplied by vector B=0, both A and B may not be zero vectors, and they only need to be vertical to satisfy the condition
So your proposition is false.

Given that B is the projective of point A (3,7,4) in the xoy plane, then vector ob^2 is equal to Given that B is a projective of point A (3,7,4) in the xoy plane, then vector ob^2 is equal to Given that B is the projective of point A (3,7,4) in the xoy plane, the vector ob^2 is equal to

B (3,7).
OB^2=9+49=58

In triangle ABC, AB=2, AC=3, vector AB•vector BC=1, then BC=

Vector BC = vector AC - vector AB, then vector A B.(vector AC - vector AB)=1,AB.AC.cosA - AB2=1, then cosA =5/6, BC =√3 from cosine theorem

Vector BC = vector AC-vector AB, then vector A B.(vector AC-vector AB)=1,AB.AC.cosA - AB2=1, then cosA =5/6, BC =√3 from cosine theorem