How to a $3D$-figure that describes all vectors making $ \text{negative dot product} $ with the vector $(1,1,1)$.
The vector $(-1,-1,-2)$ has negative do product with $(1,1,1)$ because
$(1,1,1) \cdot (-1,-1,-2)=-4<0$.
The figure is
This is just particular case. But there are many vectors having negative dot products with $(1,1,1)$.
So the angle concept is required. If the angle is $ \pi/2 < \theta<3 \pi/2$ , then $ \ \cos \theta <0$ and the dot product will be negative as $a \cdot b=|a||b| \cos \theta$.
But how to draw all the vectors that are with $ \pi/2 < \theta<3 \pi/2$ angle with $(1,1,1)$.
Help me
Best Answer
Consider a plane $x+y+z=0$ which is normal to $(1,1,1)$. This plane separates $\mathbb{R}^3$ into two parts. One is above the plane, or $x+y+z>0$, and vectors in this part forms an acute angle with the vector $(1,1,1)$. Below the plane are the vectors which form obtuse angles with $(1,1,1)$, and they all satisfies the equation $x+y+z<0$. You can also observe that if $(x,y,z)\cdot (1,1,1)<0$, then definitely $x+y+z<0$.
This works for any other vector. When a vector $(a,b,c)$ is given, vectors $(x,y,z)$ satisfying $ax+by+cz<0$ will form an obtuse angle with $(a,b,c)$, and thus the value of the inner product of those two vectors would be negative.