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This is the graph that I have in mind

My question is:

Lets say I have a unit vector a that is orthogonal to the vector p $<4,5>$. If I scale a by a scalar 't', then I have a general vector that is perpendicular to the p. This seems that the set of all the vectors that are obtained by scaling the unit vector can lie on a single line, which means the solution is a line.

But in the graph, I feel that the white region should be the solution of the vectors and region is two dimensional and not a single line.

That was my first question,

I read some answers on the internet that said that the set of all vectors orthogonal to a non zero vector lie on a single line passing through the origin. Why does the line have to pass through the origin and what does passing through the origin mean if a vector is a free vector.

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  • $\begingroup$ You seem confused about affine and vector space try to compare the definition $\endgroup$
    – LuckyS
    May 17, 2021 at 3:38

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All vectors start at the origin so a line given by all scalar multiples of a vector passes through the origin as well. Drawing the vector $(2,4)$ in the white region would not be orthogonal to $(4,5)$ since it starts at the origin, not on the line generated by scalar multiplies of $(4,5)$.

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  • $\begingroup$ The line should be scalar multiple of a which is orthogonal to the p vector? If I want to reach (2,4) by scaling a then I think it should be orthogonal. $\endgroup$
    – black mamba
    May 17, 2021 at 1:38
  • $\begingroup$ @blackmamba consider another simple case where $a = (1,0)$ is orthogonal to $p=(0,1)$. For all scalars $\alpha$ then $\alpha a = (\alpha , 0) \neq (0,1)$. You can't reach $(0,1)$ by scaling, ever, because they're not colinear. This is exactly the same situation. $\endgroup$
    – CyclotomicField
    May 17, 2021 at 1:45

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