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

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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##?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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