Given three vectors $\vec{a},\vec{b},\vec{c}$ in $\mathcal{R}^2$ plane, if each vector is antiparallel (180 degree, not 0 degree) to the sum of the rest of two vectors (e.g. $\vec{a}=-\lambda_1(\vec{b}+\vec{c}),\lambda\in\mathcal{R}^2$), what is the angle differences between them, i.e. $angle(\vec{a},\vec{b})$, $angle(\vec{b},\vec{c})$, $angle(\vec{a},\vec{c})$?
I think they should all equal to 120 degree, but have no idea how to prove it.
The next question is, suppose there are four vectors instead of three, what is the conclusion on angle differences? and how about five vectors, six vectors, and so on..?
Thanks in advance!
Best Answer
You require the three vectors to satisfy these equations: \begin{align} \vec a &= -\lambda_1(\vec b + \vec c), \tag1\\ \vec b &= -\lambda_2(\vec a + \vec c), \tag2\\ \vec c &= -\lambda_3(\vec a + \vec b). \tag3 \end{align}
Use Equations $(2)$ and $(3)$ to substitute for $\vec b$ and $\vec c$ in Equation $(1)$: $$ \vec a = -\lambda_1(-\lambda_2(\vec a + \vec c) - \lambda_3(\vec a + \vec b)). $$ Multiply through on the right and collect all the $\vec a$ terms on the left: $$ (1 - \lambda_1\lambda_2 - \lambda_1\lambda_3)\vec a = \lambda_1\lambda_2 \vec c + \lambda_1\lambda_3 \vec b. $$ Divide by the coefficient of $\vec a$: $$ \vec a = \frac{\lambda_1\lambda_2}{1 - \lambda_1\lambda_2 - \lambda_1\lambda_3} \vec c + \frac{\lambda_1\lambda_3}{1 - \lambda_1\lambda_2 - \lambda_1\lambda_3} \vec b. \tag4 $$
Since we have not yet established whether $\vec b$ and $\vec c$ are an independent set of vectors, let's consider two cases.
First consider the case where $\vec b$ and $\vec c$ are an independent set of vectors. In that case $\vec a$ can be expressed as a linear combination of $\vec b$ and $\vec c$ in only one way. Therefore the coefficients of $\vec b$ and $\vec c$ in Equation $(1)$ must match the coefficients of $\vec b$ and $\vec c$ in Equation $(4)$: $$ \frac{\lambda_1\lambda_2}{1 - \lambda_1\lambda_2 - \lambda_1\lambda_3} = \frac{\lambda_1\lambda_3}{1 - \lambda_1\lambda_2 - \lambda_1\lambda_3} = -\lambda_1. $$ This implies that $\lambda_2 = \lambda_3.$ Applying a similar procedure by substituting Equations $(1)$ and $(3)$ into Equation $(2)$, we find that $\lambda_1 = \lambda_3.$ Therefore $$ \lambda_1 = \lambda_2 = \lambda_3.$$ So we can write $\lambda_1$ in place of $\lambda_2$ and $\lambda_3$, and Equation $(4)$ then says that $$ \frac{\lambda_1^2}{1 - 2\lambda_1^2} = -\lambda_1. $$ Using the given assumption that $\lambda_1 > 0$, $$ \frac{\lambda_1}{1 - 2\lambda_1^2} = -1. $$ Multiplying by $1 - 2\lambda_1^2$ on both sides and collecting terms, $$ 2\lambda_1^2 - \lambda_1 - 1 = 0. $$ This equation has two solutions: $\lambda_1 = -\frac12,$ which we must reject because we require that $\lambda_1 > 0,$ and $\lambda_1 = 1.$ Therefore $$ \lambda_1 = \lambda_2 = \lambda_3 = 1.$$ In short, Equations $(1)$, $(2)$, and $(3)$ are all equivalent to the same equation: $$ \vec a + \vec b + \vec c = 0. \tag5$$
Now come the bad news. You can set $\vec b$ and $\vec c$ to any two linearly independent vectors in $\mathbb R^2,$ and it follows that $\vec a = -(\vec b + \vec c)$ provides a non-trivial solution to Equation $(5)$.
On the other hand, suppose $\vec b$ and $\vec c$ are not an independent set of vectors. That is, suppose $\vec b = \mu \vec c.$ Then Equation $(1)$ states that $$ \vec a = -\lambda_1(\mu + 1) \vec c, $$ that is, the three vectors $\vec a$, $\vec b$, and $\vec c$ all lie in one line. Equations $(1)$, $(2)$, and $(3)$ are again easily satisfied if $\lambda_1 = \lambda_2 = \lambda_3 = 1.$ But there are additional solutions, for example, \begin{align} \vec a &= \vec b, \\ \vec c &= -3 \vec a = -3 \vec b,\\ \lambda_1 &= \frac12, \\ \lambda_2 &= \frac12, \\ \lambda_3 &= \frac32. \end{align}
So we cannot really conclude very much about the angles between the vectors; in the case where the vectors do not all lie in one line, Equation $(5)$ imposes a few limitations such as any two angles must add up to more than $180$ degrees, but there is nothing like a requirement that any angle be $120$ degrees.
If you make the further requirement that the vectors all have equal magnitude, then the three vectors actually do have to make $120$-degree angles with each other. But if you have four equal vectors, then it is sufficient for them to be arranged in two opposite pairs, that is, $\vec a$ and $\vec c$ can be $180$ degrees apart, and $\vec b$ and $\vec d$ can be $180$ degrees apart, but $\vec a$ and $\vec b$ can make any angle you want.