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

[adrimare]

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.

Homework Equations

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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?

 

[Mark44]

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.

Homework Equations

?

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?

If you're not already doing so, draw a sketch using a three-dimensional coordinate system. It will help you get your head around these kinds of problems.

 

[adrimare]

My teacher said to forget about learning how to sketch, so that's out. Is the normal to the plane (1,0,0) then? So (0,4,0) and (0,0,-2) are both on the plane. How would I stick those points on a plane with a Cartesian equation of x=0, then? I need to get a D-value, which I can get from points, I know, but when the equation is 1x + 0y + 0z + D, would the intercepts kind of not really matter? The D-value would be 0, so the equation would be x=0, right? Or am I wrong about the normal again? Or am I supposed to be looking for the direction vector formed by the two intercepts and use that as my normal?