I have a line $L$ in $\mathbb{R}^2$ that passes through two points:
$[9;7]$ and $[1;-5]$
How do I find all unit vectors orthogonal to $L$?
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Your line has direction given by $$ \vec{u}=(9-1,7-(-5))=(8,12). $$ (ie the vector $\vec{AB}$, where $A,B$ are your two points). The set you are looking for is the set of vectors $$\vec{v}=(x,y)$$ such that $$ \vec{u}\cdot\vec{v}=8x+12y=0 $$ (ie $\vec{v}$ and $\vec{AB}$ are orthogonal) and $$ x^2+y^2=1 $$ (ie $\vec{v}$ has norm $1$). I think you can take it from here. Expect to find two answers. And let me know if you want me to expand. Note: as pointed by @zaarcis in his/her answer, there is a faster route, when one already knows that there are exactly two solutions. |
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(Lazy answer mode) Calculate vector from one point to another, then swap both coordinates of it and multiply one by $-1$ (in plane it gives orthogonal vector), finally divide this vector by its length (one answer) and multiply it by $-1$ (second answer). |
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The slope of L is $\frac{7+5}{9-1} = \frac{3}{2}$. So the slope of vectors orthogonal to L should be $\frac{-2}{3}$. So we have $v = [\sin(\operatorname{atan}(\frac{-2}{3})), \cos(\operatorname{atan}(\frac{-2}{3}))]$. |
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