Normal reaction in banked road circular motion

Fom the centre of gravity of the vehicle to the axis of the circular motion. And when banking a road, they usually have a verical axis (otherwise you would just have a sloping plane).

You should see is as: ##R_1+R_2-mg\cos 5^\circ## contribute a fraction ##\sin 5^\circ## to the actual centripetal force ##mv^2/r##.

The presentation is somewhat suggestive: ##\sum F = 0## would mean uniform linear motion, not a circular trajectory. Here ##\sum F \ne 0##. In fact, ##\sum F= {mv^2\over r}##.

 

The drawing is correct. Note how the radius is specified.

 

Centripetal force is a resultant force, not an applied force. It is the component of the resultant that is normal to the direction of travel. Where the speed is constant, that makes it the entire resultant: ΣF_other=F_centripetal.
Centrifugal force is a fictional force, but is an applied force, not a resultant: ΣF_other+F_centrifugal=0.
Since centripetal and centrifugal are equal and opposite, the equations are the same.

The vehicle is describing a circle in a horizontal plane. The centre of that motion is therefore in that plane, not down the bottom of the slope.