# Parallel Vectors

This works, right?

If we let $z$ = $(p_0, p_1, p_2)$ and $\zeta$ = $(q_0, q_1, q_2)$, and if we say $z \diamond \zeta$ defines the operation $(\frac{p_0}{q_0}, \frac{p_1}{q_1}, \frac{p_2}{q_2})$, then if each $\frac{p_i}{q_i}$ is equal for all $i$ = 0, 1, 2, then $x$ is parallel to $y$.

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What if there exists a $q_j = 0$? – apnorton Feb 22 '13 at 1:55
Need to be careful if vectors are one of the three axes (since coordinates will be zero which you can't divide). Alternately you can use the cross product; $a\times b=0$ iff $a\|b$ in ${\bf R}^3$ (this has the limitation of only working in two or three dimensions.) – anon Feb 22 '13 at 1:55
How does the image apply to your question? – apnorton Feb 22 '13 at 2:08
See my most recent edit. – Trancot Feb 22 '13 at 2:10
@Trancot I see the image was added, but I don't see how it relates to your new $\diamond$ operator... – apnorton Feb 22 '13 at 2:13

The definition provided in the book is different than the definition you have created. How can we show this? Simply provide a counter-example:

Let $\vec a = \langle 1, 0, 4 \rangle$, and $\vec b = \langle 2, 0, 8 \rangle$.

These vectors are parallel by the book's definition, as $\vec a = 2\vec b$.

However, using your $\diamond$ operator: $$\vec a \diamond \vec b = \left\langle \frac{1}{2}, \frac{0}{0}, \frac{4}{8}\right\rangle$$

We cannot conclude that $\frac{1}{2}= \frac{0}{0}= \frac{4}{8}$, as $\frac{0}{0}$ is undefined.

Thus, your definition is not equivalent.

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OK, yes. I was thinking for all points not contained in any axial plane. – Trancot Feb 22 '13 at 4:38