Comments on: Planet Genius http://involution.com/2006/12/30/planet-genius/ Tony Perrie's Weblog Sun, 15 Jan 2012 20:31:20 +0000 hourly 1 http://wordpress.org/?v=3.3.1 By: fff http://involution.com/2006/12/30/planet-genius/comment-page-1/#comment-5073 fff Wed, 03 Jan 2007 01:19:12 +0000 http://involution.com/2006/12/30/planet-genius/#comment-5073 This is in response to an article written in 2002 about finding the point on a line in 3 space closest to the origin. There is an easier way to get the solution without calculus. If "p" is the point on the line closest to the origin then the vector from the origin to p must be perpendicular to the direction vector , q = q1-q0 on the line. Let q0 = (1,2,3) , q1 = (3,5,7) , and q = q1-q0 = (2,3,4) then since p is on the line we must have: p = q0 + q*t for some scalar t and since p must be perpendicular to q we have: p . q = 0 Substituting first equation for p in second equation gives (q0+q*t) . q = 0 Solving for t gives: t = (- q0 . q) / ||q||^2 = - (1,2,3) . (2,3,4) / ||(2,3,4)||^2 = -20/29 so p = q0 - 20q/29 as in your solution. This is in response to an article written in 2002 about finding the point on a line in 3 space closest to the origin.
There is an easier way to get the solution without calculus. If “p” is the point on the line closest to the origin then the vector from the origin to p must be perpendicular to the direction vector , q = q1-q0 on the line.

Let q0 = (1,2,3) , q1 = (3,5,7) , and q = q1-q0 = (2,3,4)

then since p is on the line we must have:

p = q0 + q*t for some scalar t

and since p must be perpendicular to q we have:

p . q = 0

Substituting first equation for p in second equation gives

(q0+q*t) . q = 0

Solving for t gives:

t = (- q0 . q) / ||q||^2 =
– (1,2,3) . (2,3,4) / ||(2,3,4)||^2 =
-20/29

so p = q0 – 20q/29 as in your solution.

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