A plane passes through the straight line L: X-1 = y + 2 / 2 = Z, and then passes through the intersection of Z axis and plane: 2x-y-z + 3 = 0 to find the equation of the plane

A plane passes through the straight line L: X-1 = y + 2 / 2 = Z, and then passes through the intersection of Z axis and plane: 2x-y-z + 3 = 0 to find the equation of the plane

From the straight line X-1 = (y + 2) / 2 = Z, we can see that the normal vector of the plane is (1,2,1)
The intersection of Z-axis and plane: 2x-y-z + 3 = 0 is (0,0,3)
This plane can be expressed as x + 2Y + Z-3 = 0 by point method

It is known that x, y and Z are not zero, and 4x − 3Y − 3Z = 0x − 3Y + Z = 0

4x − 3Y − 3Z = 0 & nbsp; & nbsp; & nbsp; ① x − 3Y + Z = 0 & nbsp; & nbsp; & nbsp; ②, ① - ②: 3x-4z = 0, x = 43z, ① - ② × 4: 9y-7z = 0, y = 79z, then x: Y: z = 43z: 79z: z = 12:7:9

The linear equation passing through (1, - 1,0) and parallel to the plane x + 2y-3z + 1 = 0 and perpendicular to the straight line x + 1 / - 1 = Y-3 / 2 = Z-2 / 4

The normal vector of plane x + 2y-3z + 1 = 0 is n = (1,2, - 3),
The direction vector of the straight line (x + 1) / (- 1) = (Y-3) / 2 = (Z-2) / 4 is V1 = (- 1,2,4),
Therefore, the direction vector of the straight line is v = n × V1 = (14, - 1,4),
Therefore, the linear equation is (x-1) / 14 = (y + 1) / (- 1) = (z-0) / 4

The equation of curve y ^ 2 + Z ^ 2-2x = 0; Z = 3 projected on the x0y plane is ()
The equation of curve y ^ 2 + Z ^ 2-2x = 0; Z = 3 projected on the x0y plane is ()

When z = 3, y ^ 2 = 2x + 9 = 2 (x + 9 / 2), that is, y ^ 2 = 2x image is shifted 4.5 units to the left