The plane equation passing through the point m (- 2,7,3) and parallel to the plane x-4y + 5z-1 = 0 is?

The plane equation passing through the point m (- 2,7,3) and parallel to the plane x-4y + 5z-1 = 0 is?

Let the plane equation be x-4y + 5Z + D = 0
Point m (- 2,7,3) is substituted
-2-4*7+15+D=0
D=15
x-4y+5z+15=0

The plane equation passing through points (a, B, c) and parallel to the plane x-4y + 5z-1 = 0 is?

The plane is parallel to the known plane, so the normal vectors are the same (1, - 4,5) (i.e. the coefficients in the equation)
Let the coordinates of any point on the required plane (x, y, z),
Then 1 * (x-a) + (- 4) * (y-b) + 5 * (z-c) = 0

The solution of plane equation
Given the coordinates of a point in the plane and the equation of a straight line in the plane that does not include the point, how to solve the equation of the plane?

Through this point, any straight line intersects with a known straight line, and then the normal vectors (a, B, c) of the plane are obtained from the direction vectors of the two straight lines
Then the plane equation is x / A + Y / B + Z / C = K
Substituting known points to get k

Advanced Mathematics for plane equations!
Let a plane be perpendicular to the plane z = 0 and the equation of the plane be solved by the straight line x-2y + Z = 2x + Y-Z = - 1

If the plane is perpendicular to the plane z = 0, the plane equation can be simplified as y = ax + B
The intersection of two planes x-2y + Z = 2x + Y-Z = - 1
x=z/5
y=（-5+3z）/5
Know that (0, - 1,0) (1,2,5) is on the plane to be solved, and substitute it to obtain the plane equation as follows:
y=3x-1

Solving plane equation according to condition in Higher Mathematics
Parallel to the x-axis and passing through two points (4,0, - 2) and (5,1,7)

Let a normal vector in the plane be n = (a, B, c) (the vector sign is not marked) let P (4,0, - 2) and Q (5,1,7), then the vector PQ = (1,1,9) let a vector on the x-axis be m = (a, 0,0) (a is not equal to 0) m point multiplied by n = AA = 0, so a = 0pq point multiplied by n = B + 9C = 0, B = - 9C point formula equation a (x-4) + B (y-0) + C (Z + 2) = 0by + C

High number: find the plane equation
In three-dimensional space, a plane passes through a straight line (X-7) / 3 = (Y-8) / 4 = (Z-9) 5 and passes through a point (1,1,1). Find the equation of the plane
|x y z|
The normal vector obtained from | 1 | is: (1, - 2, 1)
|6 7 8|
How do you get this normal vector?

The direction vector of two points is (6,7,8)
|x y z|
The normal vector obtained from | 3, 4, 5 | is: (1, - 2, 1)
|6 7 8|
So the equation is (x-a) - 2 (y-b) + (z-c) = 0, which is brought into (1,1,1) to get x-2y + Z = 0

For example, x + 3 / 2 = Y-5 / 3 = Z / 1 is used to solve the plane beam equation
How to ask?

x+3-z+λ(y-5-3z)=0
You are defined by the plane bundle
Let l be represented by the equations: a1x + b1y + c1z + D1 = 0;
A2x + b2y + c3z + D2 = 0
Let's establish: a1x + b1y + c1z + D1 + λ (a2x + b2y + c3z + D2) = 0, which means a plane
This problem corresponds to: plane equations: x + 3 / 2 = Z / 1
Y-5 / 3 = Z / 1, which is the result I gave
(Note: you can also take any two: x + 3 / 2 = Y-5 / 3
Y-5 / 3 = Z / 1 is also true)

The plane equation passing through two points (4,0, - 2) and (5,1,7) and parallel to the x-axis is the process of solving

(5,1,7) - (4,0,-2) = (1,1,9)
(1,1,9) x (1,0 ,0) = (0,9,-1)
Plane equation 9 (Y - 0) + (- 1) (Z - (- 2)) = 0
9y - z = 2
If you are satisfied, please click [satisfied] in the upper right corner~

The plane π passes through points (1, - 2,1) and points (7, - 5,2) and is parallel to the x-axis
Let the plane π pass through points (1, - 2,1) and points (7, - 5,2) and be parallel to the x-axis
Because the plane is parallel to the X axis, so
Let the plane equation be
By＋Cz＋D=0
Passing through P1 (1, - 2,1) and P2 (7, - 5,2)
Namely
-2B+C+D=0
-5B+2C+D=0
And then how do you figure it out?
I think I figured it out, 3b = C, B = - D
By + 3bz-b = 0
Get rid of B
Y + 3z-1 = 0
Did I miscalculate?

Let P = (m, N, 1), a (1, - 2,1) and B (7, - 5,2) be the normal vector of plane π
The positive direction vector of X axis I = (1,0,0),
AB=(6,-3,1),
According to the meaning of the title, p * I = M = 0,
p*AB=6m-3n+1=0,
∴n=1/3,p=(0,1/3,1).
Let the normal vector of plane π be (0,1,3),
The equation of plane π is y + 2 + 3 (Z-1) = 0, that is y + 3z-1 = 0
Your answer is correct

Let the plane π pass through points (1, - 2,1) and points (7, - 5,2) and be parallel to the x-axis
Parallel to the x-axis, then a = 0
-2B+C+d=0
-5B + 2C + D = 0, get b = 0, then calculate all 0! What's wrong?

Subtract, get
3B-C=0
C=3B
-2B+3B+D=0
D=-B
thus
The equation is as follows
By+Cz+D=0
By+3Bz-B=0
The equation is as follows:
y+3z-1=0
Forget the all zero case