I am given with a line which is parameterized by their normal vectors $w \in \mathbb{R}^2$ with $| w| = 1$ and their distance $s > 0$ from origin. Now if x is any arbitrary scaler then what does $sw+xw^{\bot}$ where $w^{\bot}$ represents vector prependicular to $w$.
How to interpret the following thing physically?
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geometry
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0So in this case $|w|=1$? – 2017-02-10
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0ys sorry not to mention that. – 2017-02-10
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2$sw+xw^{\perp}$ represents the line you just described. – 2017-02-10
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0How to interpret this? I am not getting this. – 2017-02-10
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1Your line is 1) distance $s>0$ from the origin, 2) Perpendicular to $w$. There's a little bit of problem with that definition (it defines two parallel line perpendicular to $w$, but your expression defines a unique line which is one of the two). But anyway, because your line is perpendicular to $w$ it must have a point of intersection with the line with direction vector $w$, passing the origin. Lets parameterised this line by $yw$ where $y$ is a scalar. Now, your line is distance $s$ from the origin, so this point of intersection must be $sw$, because $|yw|=|w||y|=|y|$ but $|y|=s$, so $s=y$ – 2017-02-10
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1(you can also have $y=-s$ which defines another line, but as your expression tells otherwise, let's exclude this) so $sw$ is a point on your line, but you know that direction vector of this line must be $w^{\perp}$, because it must be perpendicular to $w$. So your line is parameterised as $sw+xw^{\perp}$ where $x$ is a scalar – 2017-02-10
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0great work. Thanks – 2017-02-10
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0@user160738: I suggest you combine your comments into an answer. – 2017-02-10