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

In summary: Does that describe the direction of the line? If so, then the minimum distance between the person and the object will depend on the direction of the line, and cannot be solved with just the given information. In summary, the problem is to find the minimum distance between a person at (0,0,0) and an object that hits the ground at (a,b,0) and continues propagating, but the direction of the object's trajectory is not specified by the given information.
  • #1
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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  • #2
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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  • #3
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.
 
  • #4
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 ?
 
  • #5
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##?
 

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

What is the formula for finding the minimum distance from a point to a line in 3D?

The formula for finding the minimum distance from a point to a line in 3D is:d = |(p-a) x (p-b)|/|b-a|, where p is the point, a and b are two points on the line, and d is the minimum distance.

How does the cross product play a role in finding the minimum distance from a point to a line in 3D?

The cross product is used to find the perpendicular distance from the point to the line. The distance formula involves taking the magnitude of the cross product of the vector from the point to any point on the line, and the direction vector of the line. This gives the shortest distance between the point and the line.

Can the minimum distance from a point to a line in 3D ever be negative?

No, the minimum distance from a point to a line in 3D is always positive. This is because distance is a magnitude, and cannot have a negative value.

What are the practical applications of finding the minimum distance from a point to a line in 3D?

Finding the minimum distance from a point to a line in 3D is useful in many fields, including engineering, computer graphics, and physics. It can be used to determine the shortest distance between a point and a line segment, which is important in designing structures, creating 3D models, and calculating forces in physics problems.

Is there a difference between finding the minimum distance from a point to a line in 2D vs 3D?

Yes, the formula for finding the minimum distance from a point to a line is different in 2D and 3D. In 2D, the distance can be calculated using the formula: d = |ax + by + c|/√(a² + b²), where a, b, and c are constants representing the equation of the line. In 3D, the formula involves taking the magnitude of the cross product, as mentioned in the first question.

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