# Find the Plane: Perpendicular to yz-Plane & y=-2, z=4

In summary, the student is trying to determine the equation of a perpendicular plane that has y-intercept at 4 and z-intercept at -2. The normal to the plane is perpendicular to this vector, so the plane is located at (0,1,1). The student is trying to find the equation of the plane using the y- and z-intercepts at 4 and -2, but is having trouble because the normal is not <0, 1, 1>.

## Homework Statement

Determine the Cartesian equation of the plane that is perpendicular to the yz-plane and has y-intercept 4 and z-intercept -2.

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## The Attempt at a Solution

I'm pretty sure that the normal to the plane I want to find would be (0,1,1). A y-intercept would go through (0,y,0) and a z-intercept would go through or be (0,0,z). How do I make an equation with a y and z- intercept at 4 and -2 respectively?

## Homework Statement

Determine the Cartesian equation of the plane that is perpendicular to the yz-plane and has y-intercept 4 and z-intercept -2.

?

## The Attempt at a Solution

I'm pretty sure that the normal to the plane I want to find would be (0,1,1). A y-intercept would go through (0,y,0) and a z-intercept would go through or be (0,0,z). How do I make an equation with a y and z- intercept at 4 and -2 respectively?
No, the normal to the plane is NOT <0, 1, 1>. The normal is perpendicular to this vector. The plane is perpendicular to the y-z plane. The y-intercept is 4, which means that the point (0, 4, 0) is on the plane. The z-intercept is -2, which means that what point is on the plane?