How to find the direction vector of the intersection line of two plane equations in space 2X+y=2 X-2y+z=4 |Ijk| |2 1 0| |1 -2 1| = I-2j-5k =(1 -2 -5) How do you calculate this?

The normal vectors of the equations for the two planes are:
(2,1,0) And (1,-2,1)
Then (2,1,0)×(1,-2,1)
=
|Ijk|
|2 1 0|
|1 -2 1|
= I-2j-5k
=(1 -2 -5)
That is, the direction vector of the intersection line is (1-2-5).
//
Third-order determinant algorithm:
|Ijk|
|2 1 0|
|1 -2 1|
=
|1 0|
|-2 1|* I
-
|2 0|
|1 1|* J
+
|2 1|
|1-2|*K
And
|1 0|
|-2 1|=1*1-0*(-2);
|2 0|
|1 1|=2*1-1*0;
|2 1|
|1 -2|=2*(-2)-1*1.
With regard to the calculation of determinant,

Find the plane equations of (1,2,1) and (2,-1,2) parallel to vector {3,2,1 Find the plane equations of (1,2,1) and (2,-1,2) parallel to vector {3,2,1}

Let the plane be Ax+By+Cz+D=0
The plane normal vectors are (A, B, C) and (3 2 1) Vertical
Yes 3A+2B+C=0
Point on Plane
Then A+2B+C+D=0
2A-B+2C+D=0
3 Equations 4 unknowns
AB C can be represented by D
Elimination of D on both sides of the plane equation into a general form

Let a plane pass through a point (1,0,-1) and be parallel to vector a=(2,1,1) and vector b=(1,-1,0). Let the plane pass through the point (1,0,-1) and be parallel to the vector a=(2,1,1) and the vector b=(1,-1,0).

The normal vector of a plane can be determined by a =(2,1,1) and b =(1,-1,0). n =(1,1,-3) Let (x, y, z) be the point on the plane: the equation of the plane is 1×(x-1)+1*(y-0)-3(z+1)=0=> x+y-3z-4=0

A plane equation that passes through a point (2,-1,4) and is parallel to both vectors a={-3,4,-6} and b={-2,3,-1

Provides you with precise answers
It can be known that the normal vector of this plane is a cross multiplication vector of two vectors. Let c
Then:
C = vector a x vector b =
|Ijk|
|-3 4 -6|
|-2 3 -1|
=(-4+18,12-3,-9+8)=(14,9,-1)
According to the point normal equation of the plane:
The plane equation is:14(x-2)+9(y+1)-(z-4)=0
That is,14x+9y-z-13=0
Yanbo tries to answer for you

A Problem about Vector in Senior One Mathematics Take vector OA=vector a, vector OB=vector b as parallelogram OADB, vector BM=(1/3) vector BC, vector CN=(1/3) vector CD, vector a, vector b represents vector OM, vector ON, vector MN. Steps? Please be more specific

I think there's something missing in your question. How did that C-point come about?

I think there's something missing in your question. How did that C point come about?

A vector problem 》》》》 Given a=(2x-y+1, x+y-2), b=(2,-2), when x, y is values, vectors a, b are collinear, but a is not equal to b?

The vectors a, b are collinear, i.e. the vectors a, b are parallel, so there is
-2*(2X-y+1)-2(x+y-2)=0
Get x =1/3
So as long as x =1/3, no matter what the value vector a, b will be parallel
But because but a is not equal to b, y is not equal to (-1/3), x =1/3 vector a, b is collinear

The vectors a, b are collinear, i.e. the vectors a, b are parallel, so there is
-2*(2X-y+1)-2(x+y-2)=0
Get x =1/3
So as long as x =1/3, no matter what the value vector a, b will be parallel
But because but a is not equal to b, y is not equal to (-1/3), vector a, b is collinear when x =1/3

How to find the direction vector of the intersection line of two plane equations in space 2X+y=2 X-2y+z=4 |Ijk| |2 1 0| |1 -2 1| = I-2j-5k =(1 -2 -5) How do you calculate this?