# Equation of plane perpendicular to another

1. Sep 14, 2007

### swearbynow

Find an equation of the plane that passes through point P(-4, 2, 1) and is perpendicular to the plane x+5y+2z=3

i really feel stupid with this question, i know how to do just about every other equation like this, just not with a plane perpendicular to another with a point, if anyone can just get me started i think i will be able to solve it

2. Sep 14, 2007

### arildno

Well, the normal vector to the given plane should lie within the plane you're after.

3. Sep 14, 2007

### EnumaElish

Can you write the general equation for a plane that goes through P?

4. Sep 14, 2007

### arildno

The normal vector to the given plane is parallell to (1,5,2)

It will be helpful to remember the parametric representation of a line going through the given point, and in the direction of that normal vector..

5. Sep 14, 2007

### swearbynow

nvm.

Last edited: Sep 14, 2007
6. Sep 14, 2007

### arildno

Well, that line, in its entirety, must lie in all planes that contain the given point and that are perpendicular to the given plane.

Those planes that go through the given point have the representation:
$$a(x+4)+b(y-2)+c(z-1)=0$$
where a,b,c are fixed for any one, particular plane.
What equation for a,b,c can you now form for those planes within which that line lies?

7. Sep 14, 2007

### swearbynow

ok so would x+5y+2z=8 be a correct answer

8. Sep 14, 2007

### arildno

Eeh??

When inserting the line representation into the equation for the plane, you get the equation:
$$(a+5b+2c)t=0$$
What equation must therefore a,b,c fulfill?

9. Sep 14, 2007

### swearbynow

alright im feeling stupid right now but i dont know what exactly your asking.

i was having a,b,c be the normal vector of the plane. 1,5,2. and then inserting them into the equation you had, 1(x+4) + 5(y-2) + 2(z-1)=0 and i got the answer above.

10. Sep 14, 2007

### arildno

I don't know why you deleted the line representation:
x=t-4,y=5t+2,z=2t+1

In its entirety, this line must lie in the plane(s) we are seeking:
a((t-4)+4)+b((5t+2)-2)+c((2t+1)-1)=0
In particular, it must hold for ALL values of t, showing that the constraint on the admissible planes is that their normal vectors must be perpendicular to the normal vector of the given plane, that is a+5b+2c=0.

For example, you may choose b=-1, c=1, hence a=3, yielding the admissible plane:
$$3(x+4)-(y-2)+(z-1)=0$$
that is:
$$3x-y+z=-13$$

The plane example you posted was parallell to the given plane, not perpendicular.

Last edited: Sep 14, 2007
11. Sep 14, 2007

### swearbynow

ok i think i understand now, thanks a lot

12. Sep 14, 2007

### swearbynow

ok i was lying before but now i do thats good thanks so much