- Thread starter
- #1

#### Dhamnekar Winod

##### Active member

- Nov 17, 2018

- 103

- Thread starter Dhamnekar Winod
- Start date

- Thread starter
- #1

- Nov 17, 2018

- 103

- Moderator
- #2

- Feb 7, 2012

- 2,693

The Cartesian coordinates of $P_1$ are $(\rho_1\cos\theta_1, \rho_1\sin\theta_1\cos\phi_1,\rho_1\sin\theta_1\sin\phi_1)$, and similarly for $P_2$. Take the inner product, and use the fact that $\def\bv{\mathbf{v}} \langle\bv_1,\bv_2\rangle = |\bv_1||\bv_2|\cos\gamma.$View attachment 10317

I know that $\arccos{(\cos{\phi_1}\cos{\phi_2}+\sin{\phi_1}\sin{\phi_2}\cos{(\theta_2-\theta_1)})}=\gamma$ But how can i answer the above question?

If any member knows the proof of this formula may reply to this question with correct proof.

- Thread starter
- #3

- Nov 17, 2018

- 103

In my opinion, the Cartesian co-ordinates of $P_1,P_2$ are as follows:The Cartesian coordinates of $P_1$ are $(\rho_1\cos\theta_1, \rho_1\sin\theta_1\cos\phi_1,\rho_1\sin\theta_1\sin\phi_1)$, and similarly for $P_2$. Take the inner product, and use the fact that $\def\bv{\mathbf{v}} \langle\bv_1,\bv_2\rangle = |\bv_1||\bv_2|\cos\gamma.$

So your answer differs from the above answer for Cartesian co-ordinates of $P_1, P_2$ using appropriate $\rho_i, \theta_i,\phi_i$ where i=1,2.

- Moderator
- #4

- Feb 7, 2012

- 2,693

The only difference is that I use $\theta$ and $\phi$ where you are using $\phi$ and $\theta$. If you switch the $\theta$s and $\phi$s in my hint then you should be able to prove the result.So your answer differs from the above answer for Cartesian co-ordinates of $P_1, P_2$ using appropriate $\rho_i, \theta_i,\phi_i$ where i=1,2.

- Thread starter
- #4

- Nov 17, 2018

- 103

@Opalg, Do you mean inner product=$\vec{v_1} \cdot \vec{v_2}=\left\langle v_1,v_2 \right\rangle$

- Moderator
- #4

- Feb 7, 2012

- 2,693

Yes.