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In the book of Linear Algebra by Werner Greub, at page 202 Q.11.a, it is asked that

Let $x,y,z$ be three vectors of a plane such that $x$ and $y$ are linearly interdependent and that $x+y+z = 0$.

a-) Prove that the ordered pairs $x,y; y,z; z,x$ represent the same orientation. Then show that $$\theta(x,y) + \theta(y,z) + \theta(z,x) = 2\pi$$ where the angles refer to the above orientation.

I can see that they represent the same orientation but couldn't show the followed result. I was hoping you can help me on that point.

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  • $\begingroup$ How does your text define $\theta(x,y)$? as the lesser of the two (positive) angles determined by vectors $x,y$ each based at the same point? $\endgroup$ – coffeemath Jan 6 '18 at 7:00
  • $\begingroup$ @coffeemath $\phi(x,y)$ is defined as the vector between the vector $x$ and $y$, and not necessarily positive. $\endgroup$ – Our Jan 6 '18 at 7:05
  • $\begingroup$ @coffeemath I have put a link to the of the book to the question. $\endgroup$ – Our Jan 6 '18 at 7:06
  • $\begingroup$ Don't you mean the angle between $x$ and $y$? [not the vector] since we get a sum $2 \pi$? Also should there be a "mod $2 \pi$ somewhere? $\endgroup$ – coffeemath Jan 6 '18 at 8:22
  • $\begingroup$ @coffeemath the theta is defined originally as $cos (\theta) = \frac{(x,y)}{|x||y|}$, where $|.|$ denotes the norm, and $(.,.)$ denotes the inner product. $\endgroup$ – Our Jan 6 '18 at 8:29
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Let $X=\theta(y,z)$, $Y=\theta(z,x)$, $Z=\theta(x,y)$.

$$\cos Z=\frac{x\cdot y}{|x|\cdot|y|}$$ $$\cos X=\frac{-x\cdot y-|y|^2 }{|y|\cdot|x+y|}$$ $$\cos Y=\frac{-x\cdot y-|x|^2 }{|x|\cdot|x+y|}$$

We'll need also $\sin X$ and $\sin Y$: $$\sin X=\sqrt{1-\frac{(x\cdot y)^2+2x\cdot y|y|^2+|y|^4}{|y|^2|x+y|^2}}=\frac{\sqrt{|x|^2|y|^2-(x\cdot y)^2}}{|y|\cdot|x+y|}$$

Similarly, $$\sin Y=\frac{\sqrt{|x|^2|y|^2-(x\cdot y)^2}}{|x|\cdot|x+y|}$$

Thus,

$$\begin{align}\cos(2\pi-X-Y)&=\cos(X+Y)=\cos X\cos Y-\sin X\sin Y\\ &=\frac{|x|^2|y|^2+x\cdot y(|x|^2+|y|^2)+(x\cdot y)^2}{|x|\cdot|y|\cdot|x+y|^2}-\frac{|x|^2|y|^2-(x\cdot y)^2}{|x|\cdot|y|\cdot|x+y|^2}\\ &=\frac{x\cdot y|x+y|^2}{|x|\cdot|y|\cdot|x+y|^2}\\ &=\cos Z\end{align}$$

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