13. The linear equation passing through point (1, - 3, - 2) and perpendicular to the plane x-3y + 2z-7 = 0 is

13. The linear equation passing through point (1, - 3, - 2) and perpendicular to the plane x-3y + 2z-7 = 0 is

The normal vector of the plane is (1, - 3,2), which is parallel to the straight line. The straight line passes through the point (1, - 3, - 2). The standard equation of the straight line is (x-1) / 1 = (y + 3) / 3 = (Z + 2) / 2

Hello, what is the equation of a line passing through point (1, - 1, - 2) and perpendicular to plane 2x-2y + 3Z = 0

Normal vector of plane (2. - 2.3)
Vector of straight line (2. - 2.3)
∴(x-1)/2=(y+1)/-2=(z+2)/3

The equation of a line passing through the tangent of point (1, - 1, - 2) perpendicular to the plane 2x-2y + 3Z = 0

The normal vector of plane 2x-2y + 3Z = 0 is {2, - 2,3}
The direction vector of the line is {2, - 2,3}
∵ the straight line passing through the point (1, - 1, - 2)
The linear equation is (x-1) / 2 = (y + 1) / (- 2) = (Z-3) / 3

In the space rectangular coordinate system, find the linear equation passing through the point P (1, - 1,2) and the vertical plane 2x-2y + 3Z = 1

In the space rectangular coordinate system, the linear equation passing through the point P (1, - 1,2) and perpendicular to the plane 2x-2y + 3Z = 1 is obtained
The normal vector n of plane 2x-2y + 3z-1 = 0 = {2, - 2,3}; the line is perpendicular to the plane, so it is parallel to the plane
The normal vector is the number of directions of the line, and the parameter equation of the line is as follows:
X = 1 + 2T, y = - 1-2T, z = 2 + 3T; the standard equation of straight line is (x-1) / 2 = (y + 1) / (- 2) = (Z-2) / 3 by eliminating the parameter t