Register to reply

Calculating distance between point and surface

by ImAnEngineer
Tags: distance, point, surface
Share this thread:
ImAnEngineer
#1
Sep27-09, 10:21 AM
P: 211
1. The problem statement, all variables and given/known data
Surface V: a dot x = 9 with a=(2,-3,6). Calculate the distance from O=(0,0,0) to V.


2. Relevant equations
?


3. The attempt at a solution
The shortest distance is perpendicular to V. If n is the normalvector, n dot V = 0.

I've got no clue from here. I think I need to find a proper equation for V, but I don't know how.
Phys.Org News Partner Science news on Phys.org
Wildfires and other burns play bigger role in climate change, professor finds
SR Labs research to expose BadUSB next week in Vegas
New study advances 'DNA revolution,' tells butterflies' evolutionary history
Galadirith
#2
Sep27-09, 10:53 AM
P: 109
Hey ImAnEngineer, well from the looks of it your not completely happy with the way of defining a plane in this way, so lets have a look that :

It is possible to define a plane through and known point on its surface and its normal. Know we will take the normal to be [itex]\textbf{n}[/itex] and a known point on the surface, well call it [itex]\textbf{a}[/itex]. Let us also say that any point that is also an element of the plane, ie it is a point of the surface is [itex]\textbf{r}[/itex]. Now consider the vector [itex]\textbf{r} - \textbf{a}[/itex], this vector would be parallel to the plane surface (hopefully you can see that) and if we take the dot product of that with the plane normal, then that would be 0, as the normal is perpendicular to the plane, and hence perpendicular to the vector [itex]\textbf{r} - \textbf{a}[/itex] also, which can be described in the equation:

[tex]\textbf{n}\cdot (\textbf{r} - \textbf{a}) = 0[/tex]

which can be rewritten as

[tex]\textbf{n}\cdot \textbf{r} = \textbf{n}\cdot \textbf{a}[/tex]

which you should recognize as being in the same form as the equation you wrote first (x should be a vector in case you didn't realise that, im not sure that the way you have used V is quite correct, if V is the surface as you have said then it would be the set of all elements x that satisfy the given equation, so the n dot V = 0 that you wrote wouldn't quite be correct however hopefully the above will shed light on that :D)

now what does this mean, you were nearly along the write lines with how you proposed to attempted your problem, perpendicular distance correct, and noting that you need to find the normal vector, now given the information above can you now deduce what the normal to this plane is.

Now I think that should be what you need to get started, think about how you could use the dot product and you should be aware of the other forms of the dot product, in case you arnt (although if you are doing questions like this you should really make sure you familiarise your self more with vector operations)

[tex] \textbf{a} \cdot \textbf{b} = a_{x}b_{x} + a_{y}b_{y} = |\textbf{a}||\textbf{b}|cos(\theta)[/tex]

where ax, ay etc are the components of their respective vectors and theta is the angle between them. Another way of the thinking about the dot product is that is it the projection of one vector upon another, so what happens if one of the vectors lengths was 1 ie [itex] |\textbf{a}| = 1[/itex]. Hope that helps :D
ImAnEngineer
#3
Sep27-09, 11:49 AM
P: 211
Thanks for your reply.

I didn't realise that a dot x=9 can be rewritten as 2x1 -3x2 + 6x3 = 9. So V has a normal vector: a(2,-3,6) which equals r-v. It follows that v=(-2a,3a,-6a). v lies on V, so: x1=-2a, x2=3a, x3=-6a. This can be filled in into 2x1 -3x2 + 6x3 = 9, which leaves me with a=-9/49.

From there it's easy to calculate the distance.

Galadirith
#4
Sep27-09, 01:43 PM
P: 109
Calculating distance between point and surface

Hey ImAnEngineer, I still don't think you've quite got the grasp of it. Ill try to give a better explanation, I mocked up a quick diagram which hopefully should help, and I think, especially when you are dealing with a geometrical question, trying to visualise what's gonging on really can help.

Now first I might have confused you with my explanation as I used a to represent a known point of the plane, however you also used a in your question, so instead I will now say that P represents a known point of the plane. So here's the Diagram:



Now a quick explanation, we have our 3D axis, with z going into the page. Now r represents a variable point of the plane which is equivocal to your vector x so:

[tex]
\textbf{x} = \left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right) = \textbf{r} =
\left( \begin{array}{c} x \\ y \\ z \end{array} \right)
[/tex]

Now in the diagram is the vector [tex]\vec{Pr} = \textbf{r} - \textbf{P}[/tex] (the line with arrow in the diagram represents a vector and is not pointing at something :D) and is contained within the plane, so as before:

[tex]
\textbf{n}\cdot (\textbf{r} - \textbf{P}) = 0
[/tex]

Now I need you to look at the way in which I defined to you the equation for a plane in a 3D
space:

[tex]
\textbf{n}\cdot \textbf{r} = \textbf{n}\cdot \textbf{p}
[/tex]

Your original question was

[tex]
\textbf{a}\cdot \textbf{x} = 9
[/tex]

Now you obviously saw that those two forms were the same as you then proceeded to expand the dot product of the left side of the equation. However you havn't correct identified the normal vector for the plane. the normal vector is simply a vector, and if you think about it you will see that you infact already know what the normal vector is.

I think from then on its that you not understanding the theory still.

Now what I was trying to hint at at the end of my previous post is this:

Imagine that you have a vector well call it a and another vector b. Now I also know that |a| = 1. Now let us take the dot product of these two vectors. a.b. Now think back to the various forms of the dot product I showed you in my first post, the trigonometric one and component forms.

From this we can give a qualitative explanation of what we have just done here. because |a| = 1, what we actually have is the projection of the vector b in the direction of the vector a. Now really think about how you can apply this to your problem (look for the ;D in the diagram) have you heard of a unit vector or a unit normal vector as this is what you need to know for this question. If you don't understand this perhaps talking to you teacher is the best thing to do, these are things that you do need to understand in order to answer this question. I don't want to give you to much as obviously its good for you to work it out yourself :D


Register to reply

Related Discussions
If all normals to a surface pass through a point => surface contained in a sphere Calculus & Beyond Homework 3
Please help to Find coordinates of a point formula Angle,Distance & first point known Differential Geometry 3
Calculating point X on the Golden Spiral knowing a point and a distance Calculus 0
Distance from a point on a sphere to a point on a plane... Calculus 6