Dot Product of a Non-Zero vector with a Null Vector

Question

The dot product of two vectors let us say $\vec{A}$ and $\vec{B}$ is defined as

$$\vec{A} \cdot \vec{B} \equiv AB\cos\theta,$$

where $A$ and $B$ are the magnitudes of the vectors $\vec{A}$ and $\vec{B}$ respectively. The dot product, in its definition, not only includes the magnitudes of the vectors, but also the angle between them. Why is the dot product zero, if either $\vec{A}$ or $\vec{B}$ is a null vector? If $\vec{B}$, let us say, is a null vector, then its direction is indeterminate. What can we say about the angle $\theta$ then? What would be $\theta$ i.e. the angle between $\vec{A}$ and $\vec{B}$ if $\vec{B}$ is a null vector? How can we define $\vec{A}\cdot\vec{B}$, if $\vec{B}$ is a null vector?

Apology if my question is not appropriate according to the standards, thank you.

Answer 1

If $\vec{A}$ or $\vec{B}$ is a null vector, then $A$ respectively $B$ would be zero and the lefthand side of your definition is zero as well. As everyone expects by the dot product.

Yes the angle is indeterminate but, this is not interesting in that case.

Answer 2

One way to think about it is that the definition of the dot product is made to make the dot product continuous in $\vec{A}$ and $\vec{B}$. If you fix $\vec{A}$, and let $\vec{B}\to0$, then the Squeeze Theorem from Calc I implies that $|\vec{A}\cdot\vec{B}|\to0$ (since $-1\leq\cos(\theta)\leq1$, and $B\to0$).

Answer 3

Usually by dot (inner) product we mean a symmetric, nondegenerate, bilinear form $$\mathbb{V} \times \mathbb{V} \to \mathbb{F}$$ on a vector space $\mathbb{V}$ over a field $\mathbb{F}$. (Sometimes we also require it to be positive definite, but this is not the case if there are nonzero null vectors.) The point in this statement is simply that the angle between two vectors is not part of the definition of dot product, but rather a consequence of this definition, and we can simply say that the angle between a null vector and another vector is not defined, just as we do in the familiar case when the null vector is the zero vector.

In fact, this issue is not restricted to null vectors: If the dot product has definite signature, then the quantity $|X|$ is defined to be $\sqrt{X \cdot X}$. But if it is not definite (that is, if there are nonzero null vectors), then there are also vectors $X$ such that $B(X, X) < 0$, so $X \cdot X$ is outside the usual domain of $\sqrt{\cdot}$. We thus need to introduce new conventions to interpret $|X|$, and hence the definition of the angle $\theta$ in the equation $A \cdot B = |A||B| \cos \theta$. We can do this, but this isn't very satisfying, and in my experience no one talks about concrete angles formally in this setting anyway.

Answer 4

The dot product formula between two vectors $\vec{A}=(A_x, A_y)$ and $ \vec{B} = (B_x, B_y)$ is equal to $$A_xB_x + A_yB_y = \vec{A}\cdot \vec{B}$$

And the proof that $A_xB_x+A_yB_y= |A||B|\cos(\alpha)$ with $\alpha \in [0,\pi]$ is proved using Cosine law and since the angle $\alpha$ is not well defined between a non zero vero and a zero vector therefore the equality doesn't not holds.

However the first dot product definition is well defined and suppose $\vec{A}$ is null vector and $\vec{B}$ is or is not a null vector, you have: $\vec{A}\cdot \vec{B}=0\cdot B_x+0\cdot B_y =0$.

Sometimes, by convention, since the angle is not well defined however the dot product is zero, is said that the angle is $\pi/2.$