Electronic – arduino – 10V voltage with raspberry and arduino

You likely fried the Arduino by supplying a high input voltage (you say 12V). The voltage regulator on Arduino UNO steps down voltage, e.g. 12V, by converting power to heat. Usually it work okay as long as there isn't much current being drawn from pins. I

R-Pi GPIO pins are rated at less than 20mA, and Gert Van Loos recommends using under 3mA/GPIO to remain within the design assumptions that were made for the R-Pi.

So NO do not try to supply power to the Arduino via any R-Pi GPIO pins except 5V. It is less clear how much current the 5V supply on the R-Pi GPIO socket can supply. In the absence of information, I'd assume it might be best to avoid it.

The R-Pi USB sockets are not supplied via the R-Pi's on-board power supply. Instead they come from its external power supply, and so it very much depends on the R-Pi's external power source on whether the Arduino UNO would overload anything.

Edit: R-Pi's USB and GPIO 5V are all connected to the external power supply via a 1A resettable fuse (polyfuse). That has the benefit that it protects the external power supply. It also limits the entire current drawn from the external power supply. In the case of powering an Arduino from the R-Pi, and nothing else, it should still allow the Arduino to consume up to 300mA, which should be plenty.

I'd suggest using a USB cable plugged into the RPi and the Arduino to power the Arduino, however the schematic shows the GPIO 5V has exactly the same power-supplying capability.

Be careful, some of the R-Pi external power supplies don't seem to have much 'headroom'. Where I work, students have problems sometimes because of this.

As you said, a resistive divider would be the easiest. Clamp diodes are cheap to add and provide extra protection.

^{simulate this circuit – Schematic created using CircuitLab}

If you are using different power supplies for each of the devices and want isolation between them, you can use a common optocoupler such as 4N25, PC817, etc..

^{simulate this circuit}