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The same question was asked here, but I'm confused about the answer.

I believe this question comes from the book Linear and Geometric Algebra by Alan Macdonald, where it says "Let $v_1$ and $v_2$ be given oriented lengths... as scalars $t_1$ and $t_2$ vary, $v = t_1v_1 + t_2v_2$ (eq.1) varies over the plane determined by $v_1$ and $v_2$... Let $v_0$ be another given oriented length. What does $v = v_0 + t_1v_1 + t_2v_2$ (eq.2) parameterize?". ($v_0, v_1, v_2$ are linearly independent)

Based on the definition of plane in the question, it's easy to see that $v$ in (eq.2) does not vary over the plane from (eq.1) moved by $v_0$. For example, if $t_1$ and $t_2$ are both zero, $v = v_0$ which does not lie on the plane.

I understand eq.2 represesnts a plane if we define the plane as made of end points of all $v$ in (eq.2). But we are talking about $v$, not points here. Is the question ill-formed? How should I think about this?

Nowibananatzki
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  • sum of accelerations. –  Jan 18 '20 at 22:20
  • When working in $\mathbb R^n$, it’s common to blur the distinction between points and vectors. If you don’t do this, then how can you interpret $t_1v_1+t_2v_2$ as varying over a plane in the first place? If you consider $v_0$ to be a point instead of a vector, then this is indeed a parameterization of a plane. – amd Jan 18 '20 at 22:32
  • An affine space is nothing else the a vector space after you've chosen a point as origin. – Bernard Jan 18 '20 at 22:37

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