Question

Find the equation of the vertical plane perpendicular to the y-axis and through the point (2, 3, 4). .

Answer 1

The equation of the vertical plane perpendicular to the y-axis and passing through the point (2, 3, 4) is simply x = 2. This represents a plane parallel to the yz-plane.

Step-by-step explanation:

To find the equation of the vertical plane that is perpendicular to the y-axis and passes through the point (2, 3, 4), we can use the fact that such a plane will have a constant x-coordinate, because it is not parallel to the x-axis and does not intersect it except at the given point. The plane will therefore be vertical and will have an equation where x is constant.

Since the given point has an x-coordinate of 2, the equation of the plane is simply x = 2. This equation represents a plane that is parallel to the yz-plane and crosses the x-axis at the point (2, 0, 0). It extends infinitely in the directions of the y- and z-axes.

Answer 2

Ok, the vector equation of a plane is:n∗(r−r0)=0where n is a vector normal to the plane, r is the position vector of an arbitrary point in the plane and r0 is the position vector of a given point in the plane. We need a normal vector to solve this. To get a normal vector, we use the given information that our plane is perpendicular to the y-axis. Therefore, our normal vector must be parallel to the y-axis. Any vector parallel to the y-axis will do, so I choose the unit vector <0,1,0> for simplicity. We have:<0,1,0>∗<x−2,y−3,z−4>=0giving,y=3This makes sense, since the y-coordinate must be constant, and there is no restriction on the values x,z that lie on the plane in 3-space.