If $b \cdot a = c\cdot a$ for equal-sized non-zero vectors $a, b, c$, does it follow that $c = b$?

Question

I assumed $c = a$ due to transitivity. However, I read somewhere that this is only the case iff the vectors are collinear.

I hope someone can clear this up for me.

Thanks for your attention.

Answer 1

No, it is not necessarily true. Think of your dot product as the projection of one of the vectors on the other multiplied by the latter. Then for $a \cdot b$ let's say we have a projection of $b$ on $a$ multiplied by $a$, and for $a \cdot c$ we have a projection of $c$ on $a$. Note however, that we have a "free variable" - the angle between the vectors, which can be different in these two cases and thus it is not necessarily $b = c$.

Answer 2

I assume you mean dot products?

If so, then consider $a = (1, 0, 0)$, $b = (0, 1, 0)$ and $c = (0, 0, 1)$. All three different and of length 1, yet the dot products are the same (and 0).

(There are 2-d examples as well, but that's harder to construct.)