# Orthogonal vector with fixed lenght

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?

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

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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 $\lambda = |\vec L|/|vec B|\sin\alpha$.
(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