# Describe each plane, 8x-5y=-40z

• paralian
It is also true that a single point (which you already have) and a normal vector "determine" a plane. Would you accept that as a "description" of the plane?

#### paralian

Apologies for this being a kind of stupid question...

## Homework Statement

Describe each plane

(a) $$8x-5y=-40z$$

## Homework Equations

Probably have to find the x, y, and z intercepts.

## The Attempt at a Solution

I would assume that in order to find the x intercept you set y and z equal to zero. And the same idea for the other two.

THe answer that I got was (0, 0, 0). Am I doing this wrong?

paralian said:
$$8x-5y=-40z$$I would assume that in order to find the x intercept you set y and z equal to zero. And the same idea for the other two.

THe answer that I got was (0, 0, 0). Am I doing this wrong?

Hi paralian!

No, you're completely correct …

the plane intersects all three axes at the origin, (0, 0, 0).

erm … wasn't that obvious?

Since there are an infinite number of planes that contain (0,0,0), you might want to give some other points also to "describe" the plane. Three points determine a plane.

What is required to answer such a question?

Hi Defennder!

Hint: Forget about axes … what defines a particular plane through a given point?

The dot product of the normal vector to the plane and a vector lying on the plane through that point?

Defennder said:
The dot product of the normal vector to the plane and a vector lying on the plane through that point?

Well, that will always be zero …

but what defines the plane?

(what makes one plane different from another plane?)

More specifically, what do you or your teacher mean by "describe the plane". You said you were looking for the three axis-intercepts but found that the plane passes through (0,0,0). My point before, based on what you said you were doing, was that three points "determine" a plane. Did you intend to "describe" the plane by giving the 3 intercepts? If so that is the same as giving three points. But they don't have to be the intercepts. You know one point, (0,0,0). You can find two more points in the plane by picking any two of x, y, or z to be some simple numbers (simplicity is, after all, is the whole point of the intercepts) and then solve for the third component.

It is also true that a single point (which you already have) and a normal vector "determine" a plane. Would you accept that as a "description" of the plane?