Vectors in XYZ Space with Negative Dot Product? This question is from a Linear Algebra textbook by Gilbert Strang. I got the answer, but don't understand it complete. 
Can three vectors in the $xy$-plane have $\textbf{u} ⋅ \textbf{v} < 0$ and $\textbf{v} ⋅ \textbf{w} < 0$ and $\textbf{u} ⋅ \textbf{w} < 0$? I don't know how many vectors in the $xyz$-space can have all negative dot products. (Four of those vectors in the plane would certainly be impossible...)
I don't understand why 4 vectors in the same plane can't all have negative dot products with each other? Doesn't a negative dot product just mean that the angle between the vectors is less than 90 degrees? So wouldn't you be able to have an infinite amount of vectors in the same plane that all have negative dot products with each other? Thanks for the help. 
 A: The condition asks that any pairs out of your choice of vectors should have angle between them greater than $\pi/2$. This gives us certain geometric restraints on how vectors should be organized.
Let $v_1 \in \Bbb{R}^d \setminus \{0\}$ be our first choice and $H_1 = \{ u \in \Bbb{R}^d : u \cdot v_2 \geq 0 \}$. Then our next choice $v_2$ cannot be drawn from this region. But $H_1$ is a closed half-space, which means that half of the $\Bbb{R}^d$ becomes a forbidden region. And as you keep picking vectors, this forbidden region keep expanding. The problem says that when $d = 2$, if you pick 3 vectors in this way, the forbidden region covers the entire plane $\Bbb{R}^2$ and thus you cannot pick the fourth one.
Here is another explanation which is specific to the 2D case. Assume that you have four vectors. Pick any of them and label it $v_1$, and then label the others by $v_2, v_3, v_4$ as they appear in counter-clockwise order beginning from $v_1$. Then the angles $\angle v_1v_2$, $\angle v_2v_3$, $\angle v_3v_4$ and $\angle v_4v_1$ should sum up to $2\pi$. This means that at least one of them should be $\leq \pi/2$.
A: If $\vec u \cdot \vec v < 0$, then $$\big| \vec u \big| \cdot \big| \vec v \big| \cdot \cos \theta<0,$$ where $\theta$ is the angle between $u$ and $v$, which means the angle between $\vec u$ and $\vec v$ is greater than $90^\circ$. Now apply this to your situation.
A: since $u\cdot v=|u||v|\cos \theta_{u,v}$ and $\cos(\theta)$ is negative if $\pi/2<\theta<\pi$ this can indeed work for your case by placing the three vectors at equal angular separation.
A: A negative dot product means that the angle between vectors are larger than $90^{\circ}$. The available answers here solved it in 2D space and I am going to elaborate further in 3D space. 
Lets start with a modified question: How many vectors in the $xyz$ space can have all zero dot products. The answer is straightforward -- six, one obvious vector set can be $+x$,$+y$,$+z$ and $-x$, $-y$, $-z$. 
Now lets use $+x$ as a starting vector and try to solve the original question. To ensure negative dot product with $+x$, these vectors $+y$, $-y$, $+z$, $-z$ has to shift towards $-x$, and we know that among themselves one vector has to go to get negative dot product. Also, when they shift towards $-x$, $-x$ has to go away because the angle is smaller than $90^{\circ}$ now. 
Consequently, only 4 vectors in $XYZ$ space can have all negative dot product. 
