Re: line equation in vectors

**r** = **a** + λ**c** 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 **k**X**c** 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’(**k**X**c**)

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

(**b**X**c**)X**c** 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’[(**b**X**c**)X**c**]

Re: line equation in vectors

3d version of previous post by Hartlw is wrong.

Correct Solution:

Determine **r**0 = **a** + λ0**c** where line from B intersects given line perpendicularly:

(**r**0 - **b**)∙**c** = 0 → (**a** + λ0**c** – **b**).**c** = 0, determines λ0

**r**’ = **b** + λ’(**r**0 – **b**)

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