# line equation in vectors

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• Feb 6th 2013, 07:54 AM
Hartlw
Re: line equation in vectors
r = a + λc is equation of line through a in dir of c. r is vector from origin to a point on line and λ is arbitrary parameter. A simple sketch makes this clear.
Let i, j, k be the usual 3 dimensional basis vectors and assume the given line is in i, j plane. Then kXc is normal to the line and in the plane i, j and the equation of the line through point B (location b) normal to the given line is:

r’ = b + λ’(kXc)

EDIT: whoops. line is in general 3d dir. Take a minute to revise. Sorry

(bXc)Xc is perpendicular to c and in plane of b, c. (difficult to see, actually had to ck with a couple of pencils)

r’ = b + λ’[(bXc)Xc]
• Feb 6th 2013, 01:25 PM
Hartlw
Re: line equation in vectors
3d version of previous post by Hartlw is wrong.

Correct Solution:

Determine r0 = a + λ0c where line from B intersects given line perpendicularly:
(r0 - b)∙c = 0 → (a + λ0cb).c = 0, determines λ0

r’ = b + λ’(r0 – b)

Sorry about that. In a way I’m glad, because previous solution (wrong) was almost impossible to visualize.
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