# Cartesian equation of a plane perpendicular to anothe plane with certain intercepts

## 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?

Mark44
Mentor

## 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?

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.

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?