I am getting stuck in a really easy problem in Statistical Mechanics that involves elastic collisions, it is really very shameful that I am getting stuck at such a simple thing, but from:
$$
\left\|\overrightarrow{v_{1}}\right\|^{2}+\left\|\overrightarrow{v_{2}}\right\|^{2}=\left\|\overrightarrow{u_{1}}\right\|^{2}+\left\|\overrightarrow{u_{2}}\right\|^{2}
$$
and
$$
\overrightarrow{v_{1}}+\overrightarrow{v_{2}}=\overrightarrow{u_{1}}+\overrightarrow{u_{2}}
$$
How can I get
$$
\left\|\overrightarrow{v_{2}}-\overrightarrow{v_{1}}\right\|=\left\|\overrightarrow{u_{2}}-\overrightarrow{u_{1}}\right\|
$$
I tried completing the square in the first equation like:
$$
\overrightarrow{v_{1}} \cdot \overrightarrow{v_{1}}+\overrightarrow{v_{2}} \cdot \overrightarrow{v_{2}}-2 \overrightarrow{v_{1}} \cdot \overrightarrow{v_{2}}=\left(\overrightarrow{v_{2}}-\overrightarrow{v_{1}}\right) \cdot\left(\overrightarrow{v_{2}}-\overrightarrow{v_{1}}\right)=\left\|\overrightarrow{v_{2}}-\overrightarrow{v_{1}}\right\|^{2}=\overrightarrow{u_{1}} \cdot \overrightarrow{u_{1}}+\overrightarrow{u_{2}} \cdot \overrightarrow{u_{2}}-2 \overrightarrow{v_{1}}
$$
$\overrightarrow{v_{2}}$
and then using the second equation to get:
$$
=\overrightarrow{u_{1}} \cdot \overrightarrow{u_{1}}+\overrightarrow{u_{2}} \cdot \overrightarrow{u_{2}}-2 \overrightarrow{v_{1}} \cdot\left(\overrightarrow{u_{1}}+\overrightarrow{u_{2}}-\overrightarrow{v_{1}}\right)
$$
but I cannot seem to be able to simplify this to
$$
\overrightarrow{u_{1}} \cdot \overrightarrow{u_{1}}+\overrightarrow{u_{2}} \cdot \overrightarrow{u_{2}}-2 \overrightarrow{u_{1}} \cdot \overrightarrow{u_{2}}=\left\|\overrightarrow{u_{2}}-\overrightarrow{u_{1}}\right\|^{2}
$$
Can someone help me with this? I am sure it is quite simple, but since I am stuck I am losing way too much time on this.
I’m typing in mobile so I’m ignoring all the vector signs. The problem is to show that
$$
a^{2}+b^{2}=c^{2}+d^{2}
$$
and
$$
a+b=c+d
$$
Gives you
$$
|a-b|=|c-d|
$$
From the second eq you get by squaring both sides
$$
a^{2}+b^{2}+2 a b=c^{2}+d^{2}+2 c d
$$
Using the first equation you then get
$$
a b=c d
$$
Now subtract $2 a b=2 c d$ on both sides of the first equation and you get
$$
(a-b)^{2}=(c-d)^{2}
$$
Which is the required answer. To get the vector equivalent just replace the regular product with the dot product.