The usage vector proves two parallel exercises An exercise in which the usage vector proves two sides parallel

The usage vector proves two parallel exercises An exercise in which the usage vector proves two sides parallel

1, Prove that the two normal vectors are collinear
2. Prove that the normal vector of one plane is perpendicular to the direction vector of two straight lines in the other plane.
In any case, the method of undetermined coefficients should be used to determine the normal vector of each face

Solution of mathematical proofs - Knowledge by means of vectors It is known that O is any point in the regular triangle ABC. Make vertical line from O to each side BC, CA and AB, and the vertical foot is P, Q and R respectively. Verify that AR+BP+CQ is the fixed value. Solution of mathematical proofs - Knowledge by means of vectors It is known that O is an arbitrary point in the regular triangle ABC, make vertical line from O to each side BC, CA, AB, and the vertical foot is P, Q, R respectively.

First of all, we use the expression vector AB,"·" to express the point multiplication. Because ABC is a regular triangle, we assume that the length of all three sides is 1. According to the definition of vector point multiplication, AR=·. This point multiplication represents the length of AO line segment projection in AB direction, i.e. AR.

There are five vectors, the length of any two vectors is equal to the length of the other three vectors. Verification: The sum of the five vectors is 0 Argument should be comprehensive! There are five vectors, and the length of any two vectors is equal to the length of the other three. Verification: The sum of the five vectors is 0 The demonstration should be comprehensive!

The following letters represent vectors
Abcde
|A+b|=|c+d+e|Square on both sides ==> a^2+b^2+2ab=c^2+d^2+e^2+2cd+2de+2ce
Write the other 4 expressions in turn
Bc dea
Cdeab
De abc
Ea bcd
Add the five expressions together, cancel each other, and decompose the factors.
==>(A+b+c+d+e)^2=0
So |a+b+c+d+e|=0

Can spatial vectors a, b, c not be coplanar? And certify that.

I can't.
Can be three prisms of three prisms! Although trivectors are not coplanar, they are parallel
Consider its inverse proposition:
If a, b, c are parallel, then they must be coplanar
This proposition is false
So is the original proposition.
I. e. can not

I can't.
Can be three prisms of three prisms! Although trivectors are not coplanar, they are parallel
Consider its inverse proposition:
If a, b, c are parallel, they must be coplanar
This proposition is false
So is the original proposition.
I. e. can not

No. No.
Can be three prisms of three prisms! Although trivectors are not coplanar, they are parallel
Consider its inverse proposition:
If a, b, c are parallel, then they must be coplanar
This proposition is false
So is the original proposition.
I. e. can not

Determine the shape of quadrilateral ABCD according to the following conditions, and prove that 1 Vector AD = vector BC 2 Vector AD = vector 1/3 BC 3 Vector AB = vector DC and module of vector AB = module of vector AC

1. A parallelogram because the vector AD = vector BC.
2. It can only be determined as trapezoid. Since vector AD =1/3 vector BC, AD//BC.
3. Diamond, which is proved to be a parallelogram and has AB=AC, so it is a diamond

1. A parallelogram because the vector AD = vector BC.
2. It can only be determined as trapezoid. Since vector AD =1/3 vector BC, AD//BC.
3. Diamond, it is proved to be parallelogram, and has AB=AC, so it is diamond

1. A parallelogram because the vector AD = vector BC.
2. It can only be determined as trapezoid. Since vector AD =1/3 vector BC, AD//BC.
3. Diamond, which is proved to be a parallelogram and has AB=AC, so it is a diamond.

One mathematics, vector collinear problem proof. Let o_a=e_1, ob=e_2, oc=e_3, if there are real numbers λ_1,λ_2,λ_3 such that λ_1e_1_2e_2_3e_3=0 and λ_1_2_3=0, we prove that ABC is collinear. Write down the details of the process, online and so on, thank you. One mathematics, vector collinear problem proof. Let o_a=e_1, ob=e_2, oc=e_3, if there are real numbers λ_1,λ_2,λ_3 such that λ_1e_1_2e_2_3e_3=0, and λ_1_2_3=0, we prove that ABC collinear. Write down the details of the process, online and so on, thank you.

λ1E1+λ2e2+λ3e3=0
I.e.λ1OA 2OB+(-λ1-λ2) OC=0
So λ1(OA-OC)2(OB-OC)=0
I.e.λ1CA=-λ2CB
So CA and CB are collinear, i.e. A, B and C are collinear