Proof question 120 in vector analysis

In fact, the problem is a problem of conditional probability, and the probability that is first placed in each drawer is (1-1 5)*1 8=1 10:

Write a= b=, then the problem is to find p(b|a)

p(b|a) = p(ab) p(a) (conditional probability formula).

p(ab)=(1-1 5)*(1-1 8)=7 10 where 1-1 5 refers to being in a drawer, and 1-1 8 is not in the first one.

p(a)=1-1 10=9 10 There is p(b|) compared to the twoa)=7/9

Write a= b=

p(ab)=(1-1/5)*(1-4/8)=2/5 p(a)=1-4*1/10=3/5

Hence p(b|a)=2/3

Write a= b=

p(ab)=(1-1/5)*(1-7/8)=1/10 p(a)=1-7*1/10=3/10

Hence p(b|a)=1/3

ξ∈r^n

f(q)-f(p) = f( )q-p) median value theorem.

So. f(q)-f(p)|=|∇f(ξ)q-p)| f(ξ)q-p)|

Because ||f(ξ)1

So. f(q)-f(p)|≦q-p)|

The two-dimensional Cartesian coordinate system is built, the standard orthogonal basis is selected, and the angle between the rows of the unit vectors (cos( )sin( ) and (cos( )sin( ) is -

Take the inner product of these two unit vectors and get the sedan car.

cos(α-

cos(α)sin(α)cos(β)sin(β)cos(α)cos(β)

sin(α)sin(β)

Necessity. It is known that vector b can be ,.. by vector groups a1, a2AR is represented linearly and has a unique notation.

Let b k1a1 k2a2 ......krar.

Suppose vector groups a1, a2 ,..ar, there is a set of non-zero numbers t1, t2 ,......tr such that t1a1 t2a2 ......trar＝0

So, b (k1+t1)a1 (k2+t2)a2 ......kr+tr)ar, because t1, t2 ,......tr is not all zero, so b k1a1 k2a2 ......KRAR is ...... with b (k1+t1)a1 (k2+t2)a2kr+tr)ar are two different representations that only contradict the representation.

So, vector groups a1, a2 ,..AR is linear independent.

Sufficiency. Known vector groups a1, a2 ,..AR is linear independent.

Suppose vector b is ,.. by vector groups A1, A2The expression of the linear representation of ar is not unique, let b m1a1 m2a2 ......MRAR is ...... with B N1A1 N2A2NRAR is two different representations, then (m1 n1) a1 (m2 n2) a2 ......mr nr)ar 0, where m1 n1, m2 n2 ,......mr nr is not all zero, so a1, a2 ,......AR is linearly correlated and contradicts what is known.

So vector b is ,.. by vector groups a1, a2The strap macro method of the linear representation of AR is unique.

Let k1a1+k2a2+k3a3+k4(a5-a4)=0Evidence k1 = k2 = k3 = k4 = 0

First of all, k4=0For if not, there is.

a5=-k1/k4 a1 - k2/k4 a2 - k3/k4 a3 +a4.

However, a4 can be linearly expressed by a1, a2, a3, and the above equation results in a5 linearly from a1, a2, a3, which contradicts r(a1, a2, a3, a5)=4! So k4=0

Similarly, you can export k3, k2, and k1 all of which are 0. Thus a1, a2, a3, a5-a4 are linearly independent, i.e., r(a1, a2, a3, a5-a4) = 4

Multiply a by a on both sides of a+b+c=0 at the same time, and get it.

axb + axc = 0, move the term, get axb = - axc, thus have axb = cxa

The same goes for the rest.

Proof: a b c 0

a＝－b－c

a×b＝(－b－c)×b＝－b×b－c×b∵b×b＝0，－c×b＝b×c

a×b＝b×c

The second equal sign is also provable.

Let a=(1, 2, 3, 4, 5).

b=（β1，β2，β3， β4， β5）

then b=akk= 1 0 0 0 1

Because k is not equal to 0

So r(b) = r(a).

Because 1, 2, 3, 4, 5 are linearly independent.

So r(a)=5, so r(b)=5

Thus 1, 2, 3, 4, 5 are linearly independent.

Because r(a2,a3,a4) =3

So a2, a3, a4 are linearly independent.

So a2,a3 is linearly independent (1).

And because r(a1,a2,a3) =2

So a1, a2, a3 are linearly related (2).

From (1) and (2), it is known that a1 can be represented linearly by a2, a3. (3) If the remaining volume A4 can be represented by A1, A2, A3, then the vertical macro is represented linearly by (3) A4 combustible shirt by A2, A3.

This contradicts a2, a3, a4 linear independent.

So a4 cannot be represented linearly by a1, a2, a3.

1) Sufficiency.

oa+βob=oc=(α+oc

Therefore (oa-oc)+ ob-oc)=0

Hence ca+ cb=0

Therefore, a, b, and c are collinear.

2) Necessity.

a, b, c collinear.

Thus there are real numbers s and t that are not all zeros, such that.

sac+tbc=0

That is, S(oc-oa) + t(oc-ob) = 0

So SOA+TOB=(S+T)OC

If S+T≠0, let =s (S+T) =t (S+T), have OA+ ob=oc, and + =1

If s+t=0, then ac=bc,a,b coincide, which does not conform to the meaning of the question (a,b,c are three points in the plane).In fact, if a and b coincide, the proposition is not true as long as c is not on the straight line oa.