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I am told that $v\cdot w=0$ means that the angle between the vectors $v$ and $w$ is $90$ degrees. Then I am told that the sign of $v\cdot w$ (when it isn't equal to zero) determines whether the angle between vectors $v$ and $w$ is above or below $90$ degrees; The angle is above $90$ degrees when $v\cdot w<0$ and below $90$ degrees when $v\cdot w>0$

I have a picture from the book Introduction to Linear Algebra by Gilbert Strang, but it's quite confusing for me. I don't understand how the above information is reflected in this diagram. I would like an explanation of this for me as it will help me answer a question from the section's accompanying problem set.

enter image description here

Cherry_Developer
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2 Answers2

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The dot product between two vectors $\vec v$ and $\vec w$ is given by: $$ \vec v \cdot \vec w = |\vec v||\vec w| \cos \theta $$ where $\theta$ is the angle ($0\le\theta\le \pi$) between the two vectors, so it is positive if $\cos \theta >0 \iff 0\le \theta < \pi/2$ and it is negative if $\cos \theta <0 \iff \pi/2 < \theta \le \pi$  

So, if $r$ is a straight line orthogonal to $\vec v$ and passing through its origin that divide the plane in two semiplanes, than the dot product is positive if $\vec w$ is in the same semiplane as $\vec v$ and is negative if $\vec w$ is in the other semiplane. And this is what is illustrated by the figure.

pyrrhic
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Emilio Novati
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  • I have a follow up question in regard to this. So, in order for $\overrightarrow { v } \cdot \overrightarrow { w } <0$, $\overrightarrow { v } $ would have to be in the first quadrant of the $x,y$ coordinate plane and $\overrightarrow { w } $ would have to be in the second quadrant? (or vice versa). Is this why the picture splits the angle above $90$ degrees to one half-plane and bangle below $90$ degrees to the other half plane? – Cherry_Developer Oct 02 '15 at 13:52
  • It seems to me that your figure has nothing to do with the quadrants of a reference system (there are no axis). Depending how you chose the axis the vector $\vec v$ can be in any quadrant and the kay fact that determine the sign of the dot product is the angle between $\vec v$ and $\vec w$. – Emilio Novati Oct 02 '15 at 14:13
  • What the figure illustrate is that, if $r$ is a straight line orthogonal to $\vec v$ and passing through its origin, than the dot product is positive if $\vec w$ is in the same semiplane as $\vec v$ and is negative if $\vec w$ is in the other semiplane. – Emilio Novati Oct 02 '15 at 14:17
  • The reason I am asking is because I am trying to extend this information to $3$ dimensional space to find where all the $\overrightarrow { w } \quad $'s that would make $\overrightarrow { v } \cdot \overrightarrow { w } <0$ such that $\overrightarrow { v } $ is a 3 dimensional vector. – Cherry_Developer Oct 02 '15 at 14:18
  • I've adde something to my answer. Ina $3$ dimensional space, two vectors vith common origin stay always in a plane, so you can extend the same criterion. – Emilio Novati Oct 02 '15 at 14:23
  • So the $w$ vectors would just take up half the $3$ dimensional space? I'm merely going off of the figure and what you wrote up in your explanation about it. Since by your explanation, the dot product of $v$ and $w$ is negative in one of the semi planes. – Cherry_Developer Oct 02 '15 at 14:28
  • If $W$ and $v$ are in the same semiplane the angle between them is less than $90°$ and the cos in positive. – Emilio Novati Oct 02 '15 at 14:39
  • Let us continue this discussion in chat. – Cherry_Developer Oct 02 '15 at 14:43
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You have the following characterization:

$$v\cdot w=\Vert u\Vert\Vert w \Vert\cos (\theta)$$

Where $\theta$ is the angle between $u$ and $w$, with $0\leq \theta \leq 180°$

From this, you can see for example that this angle lies between $0$ and $90°$ if and only if $\cos(\theta)>0$ if and only if $u\cdot w>0$ (since the norms $\Vert u\Vert,\Vert w \Vert$ are greater than zero)

Daniel
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