wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> easy >> Shortest Distance to a Plane
        
(Message started by: Sir Col on Mar 8^th, 2004, 3:39pm)

Title: Shortest Distance to a Plane
Post by Sir Col on Mar 8^th, 2004, 3:39pm

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.

Title: Re: Shortest Distance to a Plane
Post by Sameer on Mar 9^th, 2004, 7:01am

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.

Title: Re: Shortest Distance to a Plane
Post by towr on Mar 9^th, 2004, 7:28am

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.

Title: Re: Shortest Distance to a Plane
Post by Sameer on Mar 9^th, 2004, 7:48am

towr p is essentially the vector of point from origin and n.p/|n| is the projection of that vector on the plane!!!

Title: Re: Shortest Distance to a Plane
Post by towr on Mar 9^th, 2004, 8:06am

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.

Title: Re: Shortest Distance to a Plane
Post by Sameer on Mar 10^th, 2004, 7:00am

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.