Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given a long vector called A and a direction vector called B, how can one retrieve a position vector OC where a orthogonal line casted down to A has a fixed lenght?

enter image description here

The image above shows a 2d representation of my problem. Although, i need a solution for 3d space.

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

Let $\lambda \in \mathbb R$, $\overrightarrow{OC} = \lambda \cdot \vec B$, and $\vec L = \overrightarrow{FC}$. Denote by $\alpha$ the angle between $\vec A$ and $\vec B$. We get $\sin \alpha = \frac{|\vec L|}{|\overrightarrow{OC}|}$, so $|\overrightarrow{OC}| = |\vec L|/\sin \alpha$ and therefore $\lambda = |\vec L|/|vec B|\sin\alpha$.

HTH, AB, martini.

share|improve this answer
    
(you meant to type \vec B instead of vec B -- I would edit it for you, but there's a silly rule that edits must be at least 6 characters) –  William DeMeo Mar 7 '12 at 10:21
    
I already came up with that solution, although its currntly not working for me. I have some sign errors which seem to occur completly randomly but reading a post with the exactly same solution assures me that this is mathematical right. I was unsure though, since i never applied trigonometric functions to vectors. –  Paranaix Mar 7 '12 at 10:57
    
I finally found my problem :) . The sinus function expected the argument to be in radians but i worked with degress. Took me ages to realize :( –  Paranaix Mar 7 '12 at 11:10
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.