Intuition for cross product of vector with itself and vector with zero vector
Sophia Terry 
I'm having trouble intuiting the following two vector identities for any vector $\mathbf{v}$. I'm only asking about intuition here, and not about their proofs (which follow from definition of cross product):
$\color{green}{\mathbf{v}} \times \color{brown}{\mathbf{v}} = \mathbf{0} \tag{*}$
$\mathbf{v} \times \mathbf{0} = \mathbf{0} \tag{*}$
For (*), my intuition is that we want a vector that's perpendicular to both $\color{green}{\mathbf{v}}$ and $\color{brown}{\mathbf{v}}$. But this is the same vector, written out two times. Therefore, we want a vector that's perpendicular to just $\mathbf{v}$. Wouldn't there be infinitely many vectors that are perpendicular to any one vector? Why is it $\mathbf{0}$?
For (**), my intuition is that we want a vector that's perpendicular to both $\mathbf{v}$ and $\mathbf{0}$. Since $\mathbf{0}$ has magnitude $0$, therefore it doesn't exist "physically", so no vector can be perpendicular to it. I'm not sure about this, though.
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$\begingroup$A good way to understand this is perhaps by means of knowing what the magnitude of the cross product means. If you have two vectors $v$ and $w$, then their cross product $v \times w$ is a vector orthogonal to the plane spanned by $v$ and $w$ and with the magnitude being the area of the paralelogram that has the vectors as sides.
Now, if you get just the vector $v$ and compute $v \times v$ the magnitude of this thing should be the area of the paralelogram with $v$ and $v$ as sides. However this paralelogram is degenerate (speaking loosely, there's no paralelogram at all in truth), so that it's area should really be zero.
If you consider on the other hand $v\times 0$ this would be in magnitude the area of the paralelogram whose sides are $v$ and $0$, however, again this paralelogram is degenerate and should have no area, so that $v\times 0 $ should be really the null vector.
$\endgroup$ 1 $\begingroup$I suggest a different intuitive definition of the cross product instead: that it produces a vector perpendicular to the plane spanned by the two vectors.
Hence, for (*), there is no plane spanned and thus no output vector. For (**), again, no plane can be formed between a vector and the zero vector, so the result is zero.
nb. this is also a good definition to use if you ever go beyond 3d, where the cross product is not defined but there are similar products that talk about planes spanned by vectors.
$\endgroup$ $\begingroup$The torque about the origin, turning moment, turning effort, etc. of a force $\mathbf{F}$ acting at a point $\mathbf{d}$ is given by $\mathbf{\tau}=\mathbf{d}\times \mathbf{F}$. The direction is given by leftie-loosie, rightie-tightie and when the force is perpendicular to the displacement then the magnitude is given by $Fd$.
If you accept this then your results are almost trivial and perfectly understandable.
The first one is a stretching so no turning and the second has no force so no turning.
$\endgroup$ $\begingroup$This answer may be straying from intuition into the realm of proof, but I'll do my best to only appeal to intuition.
It might be helpful to consider the length of the cross product of two vectors, as opposed to thinking solely of its direction. The length of the cross product of two vectors is directly proportional to the length of each vector as well as the sine of the angle they span.
In the first case, the angle spanned between a vector and itself is $0$, and $\sin(0)=0$, so that we obtain a vector of length $0$. In the second case, the length of one of the vectors is $0$, so we obtain a vector of length $0$. In any inner-product space, there is only one vector of length $0$, so the results of $v\times v$ and $v\times 0$ are forced upon us by only considering length.
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