Elementary physics question related to banking on roads I am studying 12th grade physics. I encountered this concept while studying circular motion and frictional force. You can find the relevant equations here Friction on roads 
While practicing the derivation, I accidentally resolved the vectors differently (image is attached). I'm still not able to figure out what is wrong in this and I'm getting a completely different result as you can see. Which is right, N=Mgcosθ or Ncosθ=Mg? I need concept clarification. 
 A: You seem to have forgotten that there is a resultant centripetal force at work, keeping the car going along that circular trajectory. In vector terms, we should have
$$ \mathbf N + m \mathbf g = \mathbf T \,\,\, (1) $$
where $ \mathbf T $ denotes the centripetal force pointing towards the trajectory center. This equation holds for any (inertial) coordinate system we choose. In the first picture, a coordinate system with a vertical y-axis and a horizontal x-axis was chosen, and that's the smartest choice because it makes things more obvious. Breaking (1) in components would yield
$$ N \sin \theta = T $$
$$ N \cos \theta - mg = 0 $$
since $ \mathbf T $ is horizontal and since there is no acceleration along the vertical direction. It's easy to see that solving these two equations for $ T $ gives
$$ T = mg \tan \theta $$
from which you can, for example, relate the car speed to the trajectory radius.
Your coordinate system choice, however, is a bit more troublesome, although it must yield the same result if you take all forces into account, which you didn't since you forgot about centripetal force. Consider the following diagram (I apologize for it being hand-drawn):

Making the x-axis go downwards along the plane and the y-axis go upwards perpendicularly to the plane, from (1) we should have 
$$ \mathbf N \cdot \hat x + m \mathbf g \cdot \hat x = \mathbf T \cdot \hat x $$
$$ \mathbf N \cdot \hat y + m \mathbf g \cdot \hat y = \mathbf T \cdot \hat y $$
Notice that this is pretty much the same we did before on the other coordinate system, only that we're now explicitly taking scalar products to obtain the force components, which are not so obvious in our new coordinate system. By looking at the x-axis force balance, it's easy to note that $ \mathbf N \cdot \hat x = 0 $ ( $ \mathbf N $ has no x-component ) and hence, as I hope you were able to see in my diagram,
$$ mg \sin \theta = T \cos \theta$$
$$ T = mg \tan \theta $$
Which is consistent with the other coordinate's result. Turns out, $ N \cos \theta = mg $ is right, but, as we can see, $ N = mg \cos \theta $ isn't because in the second reference frame we should actually have
$$ N = mg \cos \theta + T \sin \theta $$
$$ \Rightarrow N = mg \cos \theta + mg \dfrac{ \sin ^2 \theta}{\cos \theta} = \dfrac{mg}{cos \theta} $$
$$ \Rightarrow mg = N \cos \theta $$
This illustrates how certain coordinate systems can make things considerably harder to notice; choosing one wisely makes your life much easier. Other than that, never forget to consider all forces at work in a problem -- especially if you're dealing with bodies undergoing non-uniform motion. I hope this helped clearing things out.
A: Concept clarification: You always have the right to choose any perpendicular coordinate system. 
However, in some cases, a particular coordinate system makes more sense. This occurs for example when you know the acceleration along a particular axis. In that case, you might as well select your coordinate system that coincides with that axis.
In your case, you know the horizontal and vertical accelerations. So it makes sense to use those directions in your selected coordinate system. 
You may use the slanted coordinate system, but in that case, you should identify and use the components of acceleration along those slanted axes.
A: In a stationary situation, thee car is neither levitating nor crashing into the ground.  Hence, upward and downward force must be equal.  By Newton’s third law, every force must have a reaction force.  
Your diagram is correct, only in a stationary situation, and the weight exerted on the surface must be equal to the normal force.  That is, $N=mgcos\theta$.  Read here: http://thecraftycanvas.com/library/online-learning-tools/physics-homework-helpers/incline-force-calculator-problem-solver/.
However, the book is correct for a spinning motion.  It is $Ncos\theta=mg$ in this case because centripetal force exists, and $N>mg$.  You can read more here: http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/carbank.html, which includes friction.
