Do what in the forums? The source code thing? You can use the source tag.Edit my message to see. How do you make a line perpendicular to a line mathamaticly?
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quote:Original post by joanusdmentia
yeah, |B|^2 is |B| squared is B.B
I'm just used to seeing the formula written down in text books that use |B|^2. Although they use superscript font, btw how do you do that in the forums?
HTML is supported so you can do superscript as well as subscript
Use <sub> to start a subscript block and </sub> to end a subscript block, and likewise <sup> to start a superscript block and </sup> to end it.
So |B|2.
--
Simon O'Connor
Creative Asylum Ltd
www.creative-asylum.com
[edit - pesky tags - missed a closing tag
][edited by - s1ca on January 19, 2003 1:33:36 PM]
Should be fairly easy:
given a line AB and a point P:
1)create Vector AB, say AB =
2)create a Vector perpendicular to AB the size of AB called AB* = OR <-y, x>.
3) Now make Vector PA
4) Project PA onto the newly created AB* using the dot product
5) to get D, divide that product by the length of AB
*done*
edit: oops you're working with 3D, so you can't use this method,here's another:
1)get a point on line AB, any point call it C
2)make a vector PC
3)project PC onto AB
4)you now have the distance from C to a point X ON AB that is right under P.
5)add this distance over a normalized vector of AB on to C to create X.
6)Make CX, the length of this vector is what you want.
I think that works
[edited by - Wicked Ewok on January 19, 2003 11:17:33 PM]
given a line AB and a point P:
1)create Vector AB, say AB =
2)create a Vector perpendicular to AB the size of AB called AB* = OR <-y, x>.
3) Now make Vector PA
4) Project PA onto the newly created AB* using the dot product
5) to get D, divide that product by the length of AB
*done*
edit: oops you're working with 3D, so you can't use this method,here's another:
1)get a point on line AB, any point call it C
2)make a vector PC
3)project PC onto AB
4)you now have the distance from C to a point X ON AB that is right under P.
5)add this distance over a normalized vector of AB on to C to create X.
6)Make CX, the length of this vector is what you want.
I think that works
[edited by - Wicked Ewok on January 19, 2003 11:17:33 PM]
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