# Distance from a plane to the origin question

1. Sep 17, 2012

### MartinBob

Hi, I have the following question:

Find the perpendicular distance from the origin to the plane x + 2y + 2z = 6.

My solution is incorrect but I can't find what's wrong. I did this:

I know that the distance will be a vector from the the origin to the plane, and it will be along a normal vector. Normal vector must be parallel to (1,2,2), so I solve O + (1,2,2)t satisfies plane equation to find the point where a normal to the plane will hit the origin. So 1t + 2t + 2t = 6, t = 6/5. Then the point (6/5, 12/5, 12/5) is on the plane, and it is on the normal vector from the plane towards (0,0,0) because (6/5, 12/5, 12/5) - (6/5)(1, 2, 2) = (0, 0, 0). But when I take the distance of this point, I get sqrt((36/25)+(144/25)+(144/25)) = 18/5, and this answer is not correct.

The solution in the book used projections and I understand why it worked (the answer is 2), but I can't find what went wrong in my solution. Thanks for helping.

2. Sep 17, 2012

### chiro

Hey MartinBob and welcome to the forums.

The first thing to do is to normalize your plane equation.

You have x + 2y + 2z = 6 or x + 2y + 2z - 6 = 0 or ax + by + cz + d = 0.

Now <a,b,c> constitute a vector perpendicular to the plane: we want to normalize this vector and adjust the rest of the equation.

So normalize <1,2,2> gives a length of SQRT(1^2 + 2^2 + 2^2) = SQRT(9) = 3. So by normalizing the vector we must divide our whole equation by 3 and this gives us:

(1/3)x + (2/3)y + (2/3)z + 2 = 0.

Now in normalized form if we want to find the distance from any point to the plane, we can use the fact that (2-d) = <1/3,2/3,2/3> . p where . is the dot product and p is a point.

In this form it becomes very easy to find the distance between any point let alone the origin to the plane, and by using this normalized form we plug in p = (0,0,0) to give

2-d = 0*1/3 + 0*2/3 + 0*2/3 which gives 2-d = 0 which gives d = 2.

If d = 0 then the point will lie on the plane but otherwise it will lie either below or above it depending on the sign of the answer.

3. Sep 17, 2012

### DonAntonio

No, it's not: $$\frac{6}{5}+2\,\frac{12}{5}+\,2\frac{12}{5}\neq 6$$

DonAntonio

4. Sep 17, 2012

### HallsofIvy

Staff Emeritus
No, your equation is x+ 2y+ 2z= t+ 2(2t)+ 2(2t)= 9t= 6 so t= 2/3 and the point you want is (2/3, 4/3, 4/3).

I can't make much sense of what chiro says. "Normalizing the plane" by moving the "6" to the left of the equal sign gives the same normal vector and there is no need to divide by its length. What you did was perfecty good except that you forgot the "2" coefficients in the equation of the plane.

5. Sep 17, 2012

### chiro

I'm normalizing the vector so that you can easily calculate the distance by doing a n.p = d where n is the unit normal vector for the plane and p is a point. If d = 0 then the point is on the plane.

But the thing is that this equation only works if the plane goes through the origin so you have to "shift" the term to account for this: hence why I introduced the extra term, where you "shift" the distance corresponding to what is in the plane equation.

Its basically projecting a point onto the normal vector and then correcting the distance to take into account any plane that does not go through the origin.

6. Sep 18, 2012

### HallsofIvy

Staff Emeritus
No, whether a plane, Ax+ By+ Cx= r, contains the origin or not (whether r= 0 or not), the vector <A, B, C> is perpendicular to the plane so the line through point $(x_0, y_0, z_0)$ parallel to <A, B, C>, $x= At+ x_0$, $y= Bt+ y_0$, $z= Ct+ z_0$, is the shortest distance from the point to the plane. Putting those equations into the equation of the plane, $A(At+ x_0)+ B(Bt+ y_0)+ C(Ct+ z_0)= r$ gives a single equation for t, the point on the line where the line crosses the plane. The distance between that point and the given point is the "distance between point and plane".

7. Sep 18, 2012

### chiro

It has to be normalized otherwise it won't work.

Remember you need a projection and a projection requires a unit length vector that you are projecting to.

If you don't have this and you want to do the projection approach it won't work.

The dot product of a point against a normal just gives the amount of component against the normal, but if you don't account for the distance of how far the plane is shifted away from the origin then you will screw it up and this is account by the extra term in the plane equation.

This is just basic 3D geometry 101.

8. Sep 18, 2012

### HallsofIvy

Staff Emeritus
I have no clue what you are talking about now. The perpendicular from a point to a plane has no projection onto the plane.

9. Sep 18, 2012

### DonAntonio

I'm not sure what you're saying here but the direction normal of a plane does not have to be normalized in order to calculate its

distance from the origin. In fact what the OP did was just fine, he just messed up with that "t", but for that all was correct.

DonAntonio

10. Sep 18, 2012

### DonAntonio

Well, it has a rather boring one: the zero vector on the line's point of intersection with the plane.

DonAntonio

Ps. Seriously, it is correct to say "it has no projection on the plane". I'm just joking above.

11. Sep 18, 2012

### chiro

Here is the easiest way to understand this:

Imagine a plane that goes through the origin with some normal vector.

Now if you compute <p.n>*n where ||n|| = 1 you will project a point (or more correctly a vector with its tail at that point) on to the vector n, and if you look at the length of this vector (i.e. <p,n>) then this will be the perpendicular distance from that point to the plane.

But if the plane does not go through the origin, then you need to account for this which means you consider a shifted distance with respect to its translation with respect to its normal direction vector.

A projection of projecting one vector onto another in the form of (<a,b>b)||b|| requires that you do it on a normalized vector since all projections require P^2 = P and this means you need normalized vectors.