I was revising the topic of Dot Product in Stewart's Early Transcendentals textbook and I came across something that I need further clarification on.
As seen in the screenshot below, he says "The zero vector is considered to be perpendicular to all vectors." Now this follows the statement that if the dot product is 0, then cos θ = 0.
However, if we take the dot product of a non-zero vector and the zero vector, we can't get cos θ = 0 as the norms of the 2 vectors multiplied together is 0, and since the norms are in the denominator when we try to find θ, the fraction is undefined.
Therefore, how and why do we draw the conclusion that the zero vector is considered to be perpendicular to all vectors?
Best Answer
No, you are over-reading: this statement is not being claimed to follow (be derived from) the previous sentence(s). The author is merely telling you that $\mathbf0$ is defined as perpendicular to every vector. (Side note: $\mathbf0$ is also defined as parallel to every vector.)
This is how to read the excerpt: the final line, which is boxed up, is being introduced as a definition that is motivated by the discussion preceding it. While that discussion separately considers two cases (neither $\mathbf a$ nor $\mathbf b$ are zero vs. at least one of them is zero), the definition converges the two cases, so applies to all the possibilities for $\mathbf a$ and $\mathbf b.$ (Thus, the word 'orthogonal' in Definition 7 ought to have been boldfaced, just like the two boldfaced words in the first line.)
I can only guess what you're trying to say, which sounds like this: in the equation $$0=0\times p,$$ $p$ cannot be zero and $p$ in fact equals $\frac00.$