In $\Bbb R^3$ consider an heavy point $P$ whose mass $m$ on a circumference $\Gamma$ of radius $R$, centered in the origin. Now consider that $\Gamma$ lives in the plane $$\Pi = \{( x,y,z) \in \Bbb R^3 : x=y \}$$ We have another point $Q$ same mass $m$ attached to the origin with a spring, $Q$ could move only on the line $\{y=z=0\}$ and is also attached with another string to $P$. Both strings has the same elastic constant. Find the Lagrangian of the system.

I'm a little bit confused on what coordinates I should choose. First thing I thought $(R,\theta)$ living in $\Pi$ plane and cartesian coordinates for $Q$ $$P = (R\cos\theta,R\sin\theta, 0)\\ Q = (0, 0, x)$$ I think it's wrong because here is like I'm forgetting that the plane is rotated of $\frac{\pi}4$, so an adjustment could be $$P = (R\cos\theta,R\sin\theta, 0)\\ Q = \bigg(0, \frac{\sqrt 2}{2}x, \frac{\sqrt 2}{2}x\bigg)$$ Where as the last coordinate I choose the normal of $\Pi$.

Am I on the right track or missing something? Thank you very much for your help.

Actually could be a little misleading think $R$ as a coordinates because is fixed, in terms of what I've written above nothing changes but as a clarification $R$ is fixed and the "real" coordinate is $\theta$. Without $\Gamma$ I could also think at variable $R$.

In $\Bbb R^3$ consider an heavy point $P$ whose mass $m$ on a circumference $\Gamma$ of radius $R$, centered in the origin. Now consider that $\Gamma$ lives in the plane $$\Pi = \{( x,y,z) \in \Bbb R^3 : x=y \}$$ We have another point $Q$ same mass $m$ attached to the origin with a spring, $Q$ could move only on the line $\{y=z=0\}$ and is also attached with another string to $P$. Both strings has the same elastic constant. Find the Lagrangian of the system.

I'm a little bit confused on what coordinates I should choose. First thing I thought $(R,\theta)$ living in $\Pi$ plane and cartesian coordinates for $Q$ $$P = (R\cos\theta,R\sin\theta, 0)\\ Q = (0, 0, x)$$ I think it's wrong because here is like I'm forgetting that the plane is rotated of $\frac{\pi}4$, so an adjustment could be $$P = (R\cos\theta,R\sin\theta, 0)\\ Q = \bigg(0, \frac{\sqrt 2}{2}x, \frac{\sqrt 2}{2}x\bigg)$$ Where as the last coordinate I choose the normal of $\Pi$.

Am I on the right track or missing something? Thank you very much for your help.

Actually could be a little misleading think $R$ as a coordinates because is fixed, in terms of what I've written above nothing changes but as a clarification $R$ is fixed and the "real" coordinate is $\theta$. Without $\Gamma$ I could also think at variable $R$.

Edit: Still don't know if what I've thought is correct but after a little bit of thinking spherical coordinates are a valuable choice.

Make use of the setting below

knowing that the plane $\Pi$ correspond to $\phi = \pi/4$

$$P = \bigg(\frac{\sqrt 2}{2} R\sin\theta,\frac{\sqrt 2}{2} R\sin\theta, R \cos \theta\bigg)\\ Q=(r,0,0)$$

Still wondering if the method presented above works just fine...

In $\Bbb R^3$ consider an heavy point $P$ whose mass $m$ on a circumference $\Gamma$ of radius $R$. Now consider that $\Gamma$ lives in the plane $$\Pi = \{( x,y,z) \in \Bbb R^3 : x=y \}$$ We have another point $Q$ same mass $m$ attached to the origin with a spring, $Q$ could move only on the line $\{y=z=0\}$ and is also attached with another string to $P$. Both strings has the same elastic constant. Find the Lagrangian of the system.

I'm a little bit confused on what coordinates I should choose. First thing I thought $(R,\theta)$ living in $\Pi$ plane and cartesian coordinates for $Q$ $$P = (R\cos\theta,R\sin\theta, 0)\\ Q = (0, 0, x)$$ I think it's wrong because here is like I'm forgetting that the plane is rotated of $\frac{\pi}4$, so an adjustment could be $$P = (R\cos\theta,R\sin\theta, 0)\\ Q = \bigg(0, \frac{\sqrt 2}{2}x, \frac{\sqrt 2}{2}x\bigg)$$ Where as the last coordinate I choose the normal of $\Pi$.

Am I on the right track or missing something? Thank you very much for your help.

Actually could be a little misleading think $R$ as a coordinates because is fixed, in terms of what I've written above nothing changes but as a clarification $R$ is fixed and the "real" coordinate is $\theta$. Without $\Gamma$ I could also think at variable $R$.

In $\Bbb R^3$ consider an heavy point $P$ whose mass $m$ on a circumference $\Gamma$ of radius $R$, centered in the origin. Now consider that $\Gamma$ lives in the plane $$\Pi = \{( x,y,z) \in \Bbb R^3 : x=y \}$$ We have another point $Q$ same mass $m$ attached to the origin with a spring, $Q$ could move only on the line $\{y=z=0\}$ and is also attached with another string to $P$. Both strings has the same elastic constant. Find the Lagrangian of the system.

I'm a little bit confused on what coordinates I should choose. First thing I thought $(R,\theta)$ living in $\Pi$ plane and cartesian coordinates for $Q$ $$P = (R\cos\theta,R\sin\theta, 0)\\ Q = (0, 0, x)$$ I think it's wrong because here is like I'm forgetting that the plane is rotated of $\frac{\pi}4$, so an adjustment could be $$P = (R\cos\theta,R\sin\theta, 0)\\ Q = \bigg(0, \frac{\sqrt 2}{2}x, \frac{\sqrt 2}{2}x\bigg)$$ Where as the last coordinate I choose the normal of $\Pi$.

Am I on the right track or missing something? Thank you very much for your help.

Actually could be a little misleading think $R$ as a coordinates because is fixed, in terms of what I've written above nothing changes but as a clarification $R$ is fixed and the "real" coordinate is $\theta$. Without $\Gamma$ I could also think at variable $R$.