i have 4 vectors:
- $|\vec{AC}|=|\vec{AD}|$
- $|\vec{BC}|=|\vec{BE}|$
$\angle (\vec{AC}, \vec{AD}) $ =$\angle (\vec{BC}, \vec{BE}) $
if i know the len and angle of all vectors: how do I find the Distance between D and E ?
EDIT expanding to more vectors
Given all possible vectors pairs that satisfy:
- $|\vec{xC}|=|\vec{xy}|$ Vector ends at C
- $\angle (\vec{xC}, \vec{xy}) $ =$\angle (\vec{BC}, \vec{BE}) $
how do you describe the line that forms between all points y. (line through y1, y2, and black dots in image)
Best Answer
Define $X$ as the position of $C$ and $\theta$ as the angle used
Isn't this defined by the parametric equations: $$x(t) = t + (X-t) \cos\theta$$ $$y(t) = (X-t) \sin\theta$$ solving for $t$ gives: $$ t=X-\frac{y}{\sin\theta}$$ Thus $$x=X-\frac{y}{\sin \theta}+y\frac{\cos\theta}{\sin\theta}$$ $$y=\frac{(X-x)\sin \theta}{1-\cos\theta}$$
The distance between two points is thus: $$s=\sqrt{(y-y')^2+(x-x')^2}$$ $$s=\sqrt{(\frac{\sin\theta}{1-\cos\theta})^2 (x-x')^2+(x-x')^2}$$ $$s=|x-x'|\sqrt{\frac{2}{1-\cos \theta}}$$ or if you want to put this back in terms of $t$: $$s=|t-t'|\sqrt{2(1-\cos\theta)}$$