Planet Genius
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Well, The Chupacabra has finally started a new blog. This prompted me to assemble my ragtag assortment of feeds and jigger them into a planet site for easy consumption on rainy days when I run out of milk (e.g. Tuesdays). As a homage to Chupa paying his $30 to Bob, I have named the site Planet Genius. This site combines all of the A-list bloggers that I know personally (Chupa, Nugget, Bumper, Ivo, Decibel, Doppler, and Moonwick) onto one page on the thirty of each hour just like the clappers. Stay tuned for when I RJS those star ratings and add an RSS feed to the page. |
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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.