Activity for Josh Hyattâ€
Type | On... | Excerpt | Status | Date |
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Comment | Post #285978 |
How is the constraint $T\vec{x}\wedge\vec{y}=0$ converted into a set of linear constraints? For $m=2$, one can use the general form of the area $A$ of a parallelogram defined by two vectors $\vec{u}$ and $\vec{v}$: $A=\sqrt{|\vec{u}|^2|\vec{v}|^2-(\vec{u}\cdot\vec{v})^2}$. Substituting $A=0$, this qu... (more) |
— | about 2 years ago |
Comment | Post #285978 |
If $m$ is 2, the direction constraint simplifies to a dot product which can be represented as a single linear expression. On the other hand, if $m$ is 3, the wedge product can be replaced with a cross product, but I don't see how that could be represented with fewer than 3 linear equations. I'd guess... (more) |
— | about 2 years ago |
Comment | Post #285977 |
Nice catch, thanks for pointing it out. (more) |
— | about 2 years ago |