# Set of Orthogonal Vectors in R2 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.

• You seem confused about affine and vector space try to compare the definition May 17, 2021 at 3:38

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 $$(4,5)$$.
• @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. May 17, 2021 at 1:45