Finding the Unit Normal to a Plane

[kd001]

Homework Statement

Find the unit normal to the plane 3x + 5y - z = 2. What is the distance of the plane from the origin.

Homework Equations

The Attempt at a Solution

The normal vector in component form --> (3, 5, -1)
Magnitude of the normal vector --> √35.
Divide the normal vector by its magnitude to get the unit vector-> (3/√35, 5/√35, -1/√35)
Distance between the plane and the origin --> (!EDIT!) 2 units

Did I get this right?

Thanks

 

[LCKurtz]

Homework Statement

Find the unit normal to the plane 3x + 5y - z = 2. What is the distance of the plane from the origin.

Homework Equations

The Attempt at a Solution

The normal vector in component form --> (3, 5, -1)
Magnitude of the normal vector --> √35.
Divide the normal vector by its magnitude to get the unit vector-> (3/√35, 5/√35, -1/√35)
Distance between the plane and the origin --> 15 units

Did I get this right?

Thanks

I'm afraid 15 doesn't pass the snicker test. The three intercepts are all within two units of the origin. How can the plane be that far away?

 

[kd001]

I'm afraid 15 doesn't pass the snicker test. The three intercepts are all within two units of the origin. How can the plane be that far away?

Sorry, I was looking at the wrong equation when I wrote 15. It should be 2 units shouldn't it? And did I get the unit normal right?

Thanks.

 

[D H]

You have the unit vector correct, but you still don't have the distance. What are the coordinates of the point on the plane that is closest to the origin?

 

[kd001]

Isn't the shortest distance from the origin perpendicular to the plane? Which in this case is 2 units from the equation of the plane.

From the material I've got:

"Ax + By + Cz = D, where D is the perpendicular distance to the origin."

 

[vela]

There must be more conditions on that statement you cited from your material because the equation

$$\frac{A}{2} x + \frac{B}{2} y + \frac{C}{2} z = \frac{D}{2}$$

also describes the same plane, for instance. According to that statement, the same plane would be at a distance of both D and D/2 from the origin, which doesn't make sense.

 

[kd001]

There must be more conditions on that statement you cited from your material because the equation

$$\frac{A}{2} x + \frac{B}{2} y + \frac{C}{2} z = \frac{D}{2}$$

also describes the same plane, for instance. According to that statement, the same plane would be at a distance of both D and D/2 from the origin, which doesn't make sense.

As far as I understand, the conditions imposed on the statement I cited are that the coefficients of x, y and z must be equal to the corresponding components of the unit normal to the plane. This would mean that the perpendicular distance is equal to 2/√35. Is this the right answer? I must admit I still don't have a entirely clear understanding of this and the material I've got on the topic isn't really that good. So if someone can give me an explanation as to the best way of calculating the perpendicular distance using the equation of a plane it would be very useful.

Thanks.

 

[D H]

As far as I understand, the conditions imposed on the statement I cited are that the coefficients of x, y and z must be equal to the corresponding components of the unit normal to the plane. This would mean that the perpendicular distance is equal to 2/√35. Is this the right answer?

That is the correct answer.

So, how to get that without using an ad hoc formula? (Or alternatively, derive that formula so it isn't so ad hoc.)

There is some point on the plane such that the line passing through the origin and this point is normal to the plane. The distance between the origin and the plane is equal to the distance between the origin and this point. Think of it this way: The origin, that normal point, and any other point on the plane will form a right triangle. The distance between the origin and that other point must be greater than the distance between the origin and the normal point.

The equation of a line given a point $\vec x_0$ on a line and a unit vector $\hat n$ parallel to the line is the set of points

[tex]\vec x = \vec x_0 + \lambda \hat n[/itex]

where $\lambda$ is some real number.

One way to proceed to find that normal point is to start with a point on the plane and move outwards (increasing lambda) from there. Another way is to start from the origin and move in the direction of (or opposite direction of the unit normal) until you hit the plane. In other words, find some $\lambda$ such that $\lambda \hat n$ is on the plane.

Denoting the unit vector as $\hat n = n_x\hat x + n_y\hat y + n_z\hat z$, then we are looking for a $\lambda$ such that

$$a\lambda n_x + b\lambda n_y + c\lambda n_z = d \qquad\qquad (1)$$

where $ax+by+cz=d$ is the equation of the plane. You have already found that the unit normal is $(a\hat x +b\hat y + c \hat z)/\sqrt{a^2+b^2+c^2}$ or

$$\aligned n_x &= a/\sqrt{a^2+b^2+c^2} \\ n_y &= b/\sqrt{a^2+b^2+c^2} \\ n_z &= c/\sqrt{a^2+b^2+c^2} \endaligned\qquad\qquad(2)$$

The equation for the normal point is thus

$$(a^2\lambda + b^2\lambda + c^2\lambda)/\sqrt{a^2+b^2+c+2} = d$$

or

$$\lambda = d/\sqrt{a^2+b^2+c^2} \qquad\qquad (3)$$

Combining equations (1), (2), and (3) yields the coordinates of the normal point

$$\aligned x &= \frac{ad}{a^2+b^2+c^2} \\ y &= \frac{bd}{a^2+b^2+c^2} \\ z &= \frac{cd}{a^2+b^2+c^2}$$

The distance between the origin and this point is

$$D = \sqrt{x^2+y^2+z^2} = \sqrt{a^2+b^2+c^2}\,\frac {|d|}{a^2+b^2+c^2} = \frac{|d|}{\sqrt{a^2+b^2+c^2}}$$

 

[vela]

A geometric way to think about it is to consider any vector $\vec{x}$ from the origin to a point in the plane. The distance d from the origin to the plane is the length of the projection of this vector onto the unit normal, so $d = \hat{n}\cdot\vec{x}$, which is exactly the equation of the plane where the coefficients are the components of the unit normal.