1. Jan 20, 2010

### krtica

Q: Find an equation of a plane through the point (3, -3, -2) which is parallel to the plane −2x+1y+4z=−6 in which the coefficient of x is -2.

The website asks you to format the answer equal to zero. My input is -2x+y+4z+23, but it shows wrong. I have checked my answer a couple of times, but am not sure how to approach it differently.

2. Jan 20, 2010

### tiny-tim

Welcome to PF!

Hi krtica! Welcome to PF!

That doesn't go through (3, -3, -2).

3. Jan 20, 2010

### krtica

Re: Welcome to PF!

Thanks!

Isn't the equation for a plane a(x-xi)+b(y-yi)+c(z-zi)=0? Where a,b,c are the vector components?

4. Jan 20, 2010

### tiny-tim

Yes, but if you have actual numbers, it's a lot easier to write it as ax + by + cz = …, and just fill in the number so that it fits the given point.

5. Jan 20, 2010

### krtica

By filling in the numbers, do you mean substituting the point in x,y,z? I am not quite sure exactly what you mean, my apologies.

6. Jan 20, 2010

### tiny-tim

Yes, substituting the point in ax + by + cz …

then it's bound to go through that point, isn't it?

7. Jan 20, 2010

### krtica

God! Got it.

Thank you, I really do appreciate your help.