How to get norm (magnitude) of a vector the simple way?

I just want to get the norm of

syms phi the;
c = [-cos(phi)*sin(the)^2;-sin(phi)*sin(the)^2; - cos(the)*sin(the)*cos(phi)^2 - cos(the)*sin(the)*sin(phi)^2]

norm(c,2)

isn't really simplifying anything

If I type it manually:

simplify(sqrt(cos(phi)^2*sin(the)^4+sin(phi)^2*sin(the)^4+sin(the)^2*cos(the)^2))

I get a simple answere:

(sin(the)^2)^(1/2)

[Dyuman Joshi]

Are the and phi symbolic variable?

[Niklas Kurz]

yes

[Jan]

I've tried to let c be a [1x1] to [1x4] vector with inserting some commas in the code. I do not find any combination, where norm( c ) equals (sin(the)^2)^(1/2), which is the same as abs(sin(the)).

Therefore I do not understand, how you simplify what to get this result. Of course norm does not simplify anything, because the purpose of norm() is to calculate a norm. Simplifications are done by simplify().

[David Goodmanson]

Hi Niklas,

If your last two vector elements were

- cos(the)*sin(the)*cos(phi) - cos(the)*sin(the)*sin(phi)

i.e. cos(phi) and sin(phi) not squared, then you would get what you say. But since cos(phi) and sin(phi) are squared, you don't.

[Niklas Kurz]

I'm sorry for forgetting the simicolons in c. It might have been hard for you to reproduce what I was trying to create.

Answer 1

[Nagasai Bharat]

Hi,

From the documentation of norm and simplify you could find the usage of both these functions. norm would be used to calculate the norm of a vector/matrix but not for an expression. simpify would be used in the simplification of an algebric expression.

[Niklas Kurz]

well, it actually works if u were to incorporate some assumptions:

assume(phi>0);assume(the>0); assume(phi,'real'); assume(the,'real')

then, under these conditions

simplify(norm(c))

will simplify a lot (actually >0 not necessary)