The basic theorem of collinear vector is that if a=0, then the necessary and sufficient condition of the collinear vector b and a is that there exists a unique real number λ such that b=λa. For example, if λ is equal to 0, then any two vectors are collinear. This is obviously not correct, because it does not satisfy a//b. Theorem. If I understand it in this way, then it is not a necessary and sufficient condition. What is the unique real number λ? Is λ=0 saying the particularity of collinear vector? For example, if λ=0, then b=0. In addition, if b=0, then λ=0. But it is strange because when λ=0, b=0, b (zero vector) and a (non-zero vector) are collinear, does this satisfy a//b of parallel vector?

The basic theorem of collinear vector is that if a=0, then the necessary and sufficient condition of the collinear vector b and a is that there exists a unique real number λ such that b=λa. For example, if λ is equal to 0, then any two vectors are collinear. This is obviously not correct, because it does not satisfy a//b. Theorem. If I understand it in this way, then it is not a necessary and sufficient condition. What is the unique real number λ? Is λ=0 saying the particularity of collinear vector? For example, if λ=0, then b=0. In addition, if b=0, then λ=0. But it is strange because when λ=0, b=0, b (zero vector) and a (non-zero vector) are collinear, does this satisfy a//b of parallel vector?

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As shown in the figure, in the trapezoidal ABCD, AB//DC, E was the midpoint of AD. There were four questions: 1 If AB+DC=BC, then ∠BEC=90°; 2 If ∠BEC=90°, then AB+DC=BC; 3 If BE is the bisector of ∠ABC, then ∠BEC=90°, 4 If AB+DC=BC, then CE was the bisector of ∠DCB. The number of real propositions is () A.1 B.2 C.3 D.4

After point E, make EF//CD,
AB//DC, E is the midpoint of AD,
AB//EF//CD, EF=1
2(AB+CD);
1 AB+DC=BC,
EF=1
2BC,
BEC =90°; correct;
②∵∠ BEC =90°,
EF=1
2BC,
AB+DC=BC; correct;
3 BE is the bisector of ABC,
ABE=∠FBE,
AB∥EF,
BEF=∠ABE,
BEF=∠FBE,
EF=BF,
EF=1
2BC,
BEC =90°; correct;
4 AB+DC=BC,
EF=CF=1
2BC,
FEC=∠FCE,
EF∥CD,
FEC=∠DCE,
DCE=∠FCE,
I.e. CE is the bisector of ∠DCB, correct.
Therefore, D.

For example, ABCD is trapezoidal, AB‖CD, and AB=2CD, M, N are the midpoint of DC and AB, respectively, known as →AB=a,→AD=b, Trial a and b represent →DC,→BC and →MN respectively.

AB‖CD, AB=2CD,→AB=a
DC=-a/2
AC=→AD+→DC=b-a/2
BC=→BA+→AC=b-3a/2
MN=→MC+→CA+→AN=→DC/2+→CA+→AB/2=a/4-b/2

Given the coordinates of vectors a, b as (-1,3) and (4,-12), find the relation between a and b?

There are two ways to judge:
-1*(-12)=3*4 So parallel i.e. for any two vectors a=(x1, y1) b=(x2, y2)
When x1*y2=x2*y1, the two vectors are parallel
From the coordinates of a. b vector, we can obtain that b=4a conforms to a=λb, so it is parallel.

There are two ways to judge:
-1*(-12)=3*4 So parallel i.e. for any two vectors a=(x1, y1) b=(x2, y2)
When x1*y2=x2*y1, the two vectors are parallel
From a.b vector coordinates, we can get that b=4a conforms to a=λb, so parallel

There are two ways to judge:
-1*(-12)=3*4 So parallel i.e. for any two vectors a=(x1, y1) b=(x2, y2)
When x1*y2=x2*y1, the two vectors are parallel
From the a.b vector coordinates, we can get that b=4a conforms to a=λb, so parallel

Find the cosine of the angle between vector a (3,-4) and vector b (5,12) RT

From a =(3,-4) OA =5,
By b=(5,1) B=13,
AB=√[(5-3)2+(12+4)2]=√260.
By cosine theorem:
△ AOB: cos∠ AOB=(52+13 2-260)/2×5×13
=-33/65.

Vector a=(3,4) Vector b is opposite to vector a and vector b|=3 finds vector b coordinate

Method 1: The unit vector opposite to vector a is (-3/5,-4/5), so that vector b=t (-3/5,-4/5) t >0. Since |vector b|=3, t=3, vector b=(-9/5,-12/5)

Method 1: The unit vector opposite to vector a is (-3/5,-4/5), so that vector b=t (-3/5,-4/5) t >0, because |vector b|=3, t=3, so vector b=(-9/5,-12/5)