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Topic: Shortest Distance to a Plane (Read 1437 times) |
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Sir Col
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Shortest Distance to a Plane
« on: Mar 8th, 2004, 3:39pm » |
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Given a plane [prod] with normal vector, n=(4,2,4) and the point P (3,2,-1), which lies on the plane, find the shortest distance from the origin to the plane.
If the normal vector, n=(a,b,c) and P, which lies on the plane, is (x,y,z), find the shortest distance from the origin to the plane.
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Sameer
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Re: Shortest Distance to a Plane
« Reply #1 on: Mar 9th, 2004, 7:01am » |
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For first part, distance = 12/6 = 2 units.
Generic form, distance from origin = absolute(ax+by+cz)/sqrt(a2+B2+c2)
Can be obtained by getting the equation of plane and then considering the projection of vector of the given point (from the origin) on the plane.
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| « Last Edit: Mar 9th, 2004, 7:03am by Sameer » |
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towr
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Re: Shortest Distance to a Plane
« Reply #2 on: Mar 9th, 2004, 7:28am » |
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another way to get the same answer is to use n.p/|n|
n.p/(|n||p|) = cos([alpha]) where [alpha] is the angle between the normal vector (n) of the plane and the vector (p) from the origin to point P
|p| is the distance from the origin to P, and so |p| cos([alpha]) = np/|n| gives the distance to the plane.
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Sameer
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Re: Shortest Distance to a Plane
« Reply #3 on: Mar 9th, 2004, 7:48am » |
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towr p is essentially the vector of point from origin and n.p/|n| is the projection of that vector on the plane!!!
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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towr
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Re: Shortest Distance to a Plane
« Reply #4 on: Mar 9th, 2004, 8:06am » |
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No, it's not a projection on the plane, that would be |p|sin[alpha]. It's a projection on a line through the origin with direction vector n though, if that's what you mean.
Anyway, I just want to avoid having to explicitly make the equation for the plane first, since that's just unnecessary extra work.
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| « Last Edit: Mar 9th, 2004, 8:13am by towr » |
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Sameer
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Re: Shortest Distance to a Plane
« Reply #5 on: Mar 10th, 2004, 7:00am » |
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Well when I say "plane" I mean the normal vector as it essentially denotes what is a plane. So yes it is the projection of the vector p (whose tail on origin) on to the normal vector giving you the shortest distance. As a matter of fact.
If n=(a,b,c) and P=(x,y,z) and let us find the distance from a given point Q(p,q,r)
Equation of plane
(X-x)a+(Y-y)b+(Z-z)c = 0 (which comes from n dot (r-r0) = 0 where r = (X,Y,Z) generic vector and r0 is the given point (x,y,z) )
The vector QP = (p-x,q-y,r-z)
Projection of this vector over the normal vector gives the shortest distance i.e. QP cos(alpha)
Minimum distance = abs(QP dot n)/|n|
=abs((p-x,q-y,r-z) dot (a,b,c))/sqrt(a2+b2+c2)
=abs( (p-x)a+(q-y)b+(r-z)c)/sqrt(a2+b2+c2)
The numerator is essentially the equation of plane evaluated at the point Q, which that is why helps us in getting the distance faster.
Of course with Q as origin this falls to the simple case.
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
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