Draw a 3-dimensional vector $v$ by an arrow and show where to find all $w$'s satisfying $ v \cdot w<0$

Question

If $v \cdot w$ is negative, what can be the interpretation about the angle between $v$ and $w$?

Draw a 3-dimensional vector $v$ by an arrow and show where to find all $w$'s satisfying $ v \cdot w<0$.

Answer:

Since $ v \cdot w $ is negative, the angle between $v$ and $w$ is an obtuse angle.

For if $ \theta$ be the angle between $v$ and $w$, then

$v \cdot w=|v||w| \cos \theta \\ \Rightarrow \cos \theta=\frac{v \cdot w}{|v||w|}<0 , \ (\because v \cdot \ is \ negative) \\ \Rightarrow \frac{\pi}{2}< \theta< \frac{3 \pi}{2} $.

Next part,

If I take the vector $ v$ in the first quadrant and the vector $w$ in the above range $\frac{\pi}{2}< \theta< \frac{3 \pi}{2} $ is then the dot product of $v$ and $w$ will be negative.

But I can not draw the 3-dimensional case.

Help me drawing the sketch.

Answer 1

Attached a plot showing in red the vector $\vec v = (1,1,1)$ and in yellow the semi-space corresponding to $\vec v \cdot \vec p \le 0$ with $\vec p = (x,y,z)$

v = {1, 1, 1}; p = {x, y, z}; o = {0, 0, 0}; gr1 = RegionPlot3D[v.p < 0, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, PlotStyle -> Directive[Yellow, Opacity[0.5]], Mesh -> None]; gr2 = Graphics3D[ {Red, Thick, Arrow[{o, v}]}]; Show[gr1, gr2]

Answer 2

On the plane (2D) points that have negative dot product to a given vector are all those whose angle to the original vector is obtuse.

This property holds for the space (3D). And holds for any other dimension (ND)

In terms of visualization, in 2D you can think that given a vector $v$, we can cut the plane with a line orthogonal to $v$ that passes the origin splitting the plane into two regions. One of which contains the original vector $v$. The points with negative dot product are precisely the other region.

In 3D it is the same except instead of cutting the space with a line orthogonal to $v$ you cut it with an orthogonal plane that goes through the origin.