Exercise 1.2.30

Question

Can three vectors in the $x y$ plane have $u \cdot v<0$ and $v \cdot w<0$ and $u \cdot w<0$ ? I don't know how many vectors in $x y z$ space can have all negative dot products. (Four of those vectors in the plane would certainly be impossible ...).

Prakash Anand · Accepted Answer

Three vectors in the plane could make angles greater than $90^{\circ}$ with each other: for example $(1,0),(-1,4),(-1,-4)$. Four vectors could not do this ( $360^{\circ}$ total angle). How many can do this in $\mathbf{R}^{3}$ or $\mathbf{R}^{n}$ ? Ben Harris and Greg Marks showed me that the answer is $n+1$. The vectors from the center of a regular simplex in $\mathbf{R}^{n}$ to its $n+1$ vertices all have negative dot products. If $n+2$ vectors in $\mathbf{R}^{n}$ had negative dot products, project them onto the plane orthogonal to the last one. Now you have $n+1$ vectors in $\mathbf{R}^{n-1}$ with negative dot products. Keep going to 4 vectors in $\mathbf{R}^{2}:$ no way!

Exercise 1.2.30

Answers

Comments

Exercise 1.2.30

Answers

Comments

Add answer