Particle distance from a given plane

[Combinatus]

Distance of parabola from a given plane

Homework Statement

A particle has the position $$P(t) = (1,t,t^2)$$ at time t. Determine a formula for its distance to the plane 3x+2y+z=1. For what value of t is the particle closest to the plane?

Homework Equations

The Attempt at a Solution

I'm not perfectly sure why my solution doesn't seem to work. It gets pretty close to the solution, and I don't think I've messed up the arithmetic along the way, so there should be some lack of conceptual understanding that causes it.

Anyway, first off, a normal vector to the plane is $$n=(3,2,1)$$.

Choosing an arbitrary point in the plane, say, $$O = (0,0,1)$$, it is realized that $$\vec{OP} = \vec{v} + sn$$ where v is a vector parallel to the plane, and sn, for some value of s, is the distance from a point on the plane to the point P on the parabola that describes the path of the particle.

We then get $$\vec{v} = \vec{OP} - sn = (1,t,t^2-1) - (3s,2s,s) = (1-3s,t-2s,t^2-1-s)$$. Since $$\vec{v}$$ is parallel to the plane, its coordinates satisfy the equation of the plane. Thus, $$3(1-3s) + 2(t-2s) + t^2 - 1 - s = 1$$, which can be simplified to $$s = (1/14) \cdot (t^2 + 2t + 1)$$.

Using this value of s in $$sn$$ yields that the vector from the plane to the point P is $$(1/14) \cdot (3t^2 + 6t + 3, 2t^2 + 4t + 2, t^2 + 2t + 1)$$. The absolute value of this vector is then, after simplifying, $$(t+1)^2$$, which should be the sought distance.


Turns out that it isn't. The actual distance is supposedly given by $$\frac{t^2 + 2t + 2}{\sqrt{14}}$$. I suppose I could just try to project $$\vec{OP}$$ onto n straight away, but I don't see what exactly it is that makes my aforementioned approach fail.

I'm inclined to guess that it has something to do with the parameter s being fixed on some particular value, thus not making it possible to write it in terms of another parameter.

Ideas are welcome!

 

[lanedance]

i'm not totally sure what you did... but the shortest distance to the plane will be orthogonal to the plane (ie in the direction of the plane normal - why?)

so i would find the equation of the line thourgh the point in the direction of the normal and find where it intersects the plane

 

[Combinatus]

i'm not totally sure what you did... but the shortest distance to the plane will be orthogonal to the plane (ie in the direction of the plane normal - why?)

so i would find the equation of the line thourgh the point in the direction of the normal and find where it intersects the plane

Hello! :)

Yes, I used the realization that the shortest distance is orthogonal to the plane.

Basically, I drew a right triangle using the point O=(0,0,1) on the plane, the point on the plane that is just below some arbitrary point on the curve (call it P0 or something), and the point P(t) on the parabola. Then, the vector from P0 to P(t) should be equal to the s*n, where n=(3,2,1) is the normal from the plane and s is a parameter.

Also, $$\vec{OP} = \vec{OP_{0}} + \vec{P_{0} P(t)} = \vec{OP_{0}} + s \vec{n}$$

The sought distance, then, should be the lengh of the projection of the vector OP on the normal. However, my attempt to find this projection (mentioned above) apparently failed.Edit: Fixed my vector equation.

 

[lanedance]

sorry, my computer is struggling to see tex today, but i think i understand what you're trying to do

so say
O - arbitrary point on plane
P₀ - point on curve

so let
p = P₀ - O

the length projection of this vector onto the normal vector, should be your distance, so as you know n,

why not find |n|, and consider a dot product?

 

[LCKurtz]

You have $\vec v = \langle 1,t,t^2-1\rangle$ a vector from a point on the curve to a point on the plane. All you need is the absolute value of the projection of this on your unit normal vector:

$$d = |\vec v \cdot \hat n| = |\langle 1,t,t^2-1\rangle\cdot\frac 1{\sqrt{14}}\langle 3,2,1\rangle|$$

 

[Combinatus]

Thanks for your help; it worked out. LCKurtz, I don't understand how you projected it straight away on the unit normal vector. I think I did something equivalent though.

I guess I'll write it, too! The projection of $$\vec{OP}$$ ($$O=(0, 0, 1)$$ being a the point on the plane; $$P=(1, t, t^2)$$ being the point on the curve) is $$\lambda \vec{n}$$. (With $$\vec{n} = (3, 2, 1)$$ being the normal vector of the plane.) Then, using the dot product as lanedance said, we get $$(\vec{OP} - \lambda \vec{n}) \cdot \vec{n} = 0 \Leftrightarrow \vec{OP} \cdot \vec{n} - \lambda \vec{n} \cdot \vec{n} = 0 \Leftrightarrow \lambda = \frac{\vec{OP} \cdot \vec{n}}{|n|^2} = \frac{(1,t,t^2-1) \cdot \vec{n}}{14} = \frac{t^2 + 2t + 2}{14}$$

Thus, the sought distance $$d = |\lambda \vec{n}| = \lambda |\vec{n}| = \frac{t^2 + 2t +2}{\sqrt{14}}$$

Finally, $$t=-1 \Rightarrow d = \frac{1}{\sqrt{14}}$$ is the shortest distance.

Fin.

 

[LCKurtz]

Thanks for your help; it worked out. LCKurtz, I don't understand how you projected it straight away on the unit normal vector.

Suppose you are given a plane and a point P not on the plane. Take any point Q in the plane, and consider the vector V = QP. Now imagine looking at the plane from the edge of the plane, so it looks like a line and you see the vector V going from the point Q to P and the unit normal vector sticking up from the plane. Note that the projection of V on N will be the same no matter where Q is in the plane. That is why you can project it immediately.