Min. distance from a point to a line in 3d

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Homework Help Overview

The discussion revolves around finding the minimum distance from a point in 3D space, specifically from a person located at (0,0,0), to a line defined by an object that hits the ground at (a,b,0) and continues in a direction determined by angles θ and φ.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of a distance formula involving cross products and direction vectors. There is an exploration of the correct formulation of the direction vector and its implications on the distance calculation. Questions arise regarding the correctness of the assumptions made about the angles θ and φ and their role in defining the line.

Discussion Status

Participants are actively engaging in clarifying the definitions and relationships between the variables involved. Some suggest re-evaluating the direction vector and its components, while others question the validity of the derived distance formula based on specific conditions. There is no explicit consensus yet, but the dialogue is productive in examining the underlying concepts.

Contextual Notes

There is a mention of a potential misunderstanding regarding the meaning of the angles θ and φ in relation to the line's direction. The original poster's assumptions about the distance formula and its components are under scrutiny, indicating a need for further clarification on the problem setup.

mars187
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Homework Statement


There's a person on the ground in (0,0,0). An object comes from the sky and hits the ground at (a,b,0), but continues propagating, that is, it does not stop. I need to find the minimum distance between the person and the object.

All we know is that the person is in (0,0,0), the object hits the surface on (a,b,0) and the object comes with ##\theta## and ##\phi## (see this figure).

Homework Equations



I'm using the following to try to solve this:

$$d = \dfrac{|\vec{M_0}\vec{M_1} \times \vec{s}|}{|\vec{s}|}$$

where ##M_1## is a point in the line, $M_0$ is the point (0,0,0) and #\vec{s}# is the directing vector.

The Attempt at a Solution


[/B]
So, what I did was apply the equation of ##d##, with ##M_0 = (0,0,0)##, ##M_1 = (a,b,0)## and ##\vec{s} = (a+\sin(\theta)*\cos(\phi); b + \sin (\phi) \sin (\theta); \cos (\theta))##. It comes:$$d = \sqrt{(a^2+b^2)*\cos^2 (\theta) + a^2*\sin^2(\theta)*\sin^2(\phi)+b^2*\sin^2(\theta)*\cos^2(\phi)-2ab*\sin^2(\theta)\cos(\phi)*\sin(\phi)}$$

This is not correct, because for ##\phi = pi/2##, d should be ##d = (a^2+b^2)\cos (\theta)##..

Where have I gone wrong?
 
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Hello mars, :welcome:

Check what is meant with a direction vector. The line is described by ##\vec M_1+\alpha \vec s## with ##\ \alpha \ ## a real number; so I think your direction vector should not contain ##a## and ##b##.
 
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Are you saying that ##\vec{s} = (\sin(\theta)*\cos(\phi); \sin(\phi)*\sin(\theta); \cos(\theta))##? I agree, and that's what I used in the denominator (unknowingly).

Instead of using the equation for ##d## that I showed, one could use the one that's in the following link: http://mathworld.wolfram.com/Point-LineDistance3-Dimensional.html

In there, the denominator is |\bold{x_2} - \bold{x_1}|, which would be what you're saying. However, the result is still wrong. I've made a mistake somewhere else.
 
mars187 said:
that's what I used in the denominator (unknowingly).
You also used it in the numerator !

mars187 said:
However, the result is still wrong.
How do you know it's wrong ? Because it doesn't satisfy the requirement? (if I assume you meant ##\ d = \sqrt{a^2+b^2}\; \cos\theta\ ## ) you mentioned. Then perhaps we should check if the requirement is correct ?
 
mars187 said:

Homework Statement


There's a person on the ground in (0,0,0). An object comes from the sky and hits the ground at (a,b,0), but continues propagating, that is, it does not stop. I need to find the minimum distance between the person and the object.

All we know is that the person is in (0,0,0), the object hits the surface on (a,b,0) and the object comes with ##\theta## and ##\phi## (see this figure).

What does it mean when you say a line "comes with" ## \theta## and ##\phi##?
 

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