# Probability of a vector lying in an interval

Suppose we have a vector \$vec{v}\$ with constant length with direction specified by \$theta\$ in a random direction w.r.t the \$x\$ axis.

How can I compute the probability of the \$x\$ component of \$vec{v}\$ lying in the range \$v_{x}\$ to \$v_{x} + mathop{dx}\$

By working and looking online, I have been able to figure out the following:

Let \$|v|\$ denote the magnitude of the vector \$v\$. Then,

\$v_{x}\$ = \$|v|cos(theta)\$

We have \$P(v_{x} leq u) = P(vcos(theta) leq u) = P(vcos(theta) leq u, pi leq theta leq 2pi) + P(vcos(theta) leq u, pi leq theta leq 2pi)\$.

Let the variable \$w = u/v\$. Then, our probability \$P(v_{x} leq u)\$ equals

\$P(theta geq arccos(w), 0 leq theta leq pi) + P(theta leq arccos(w), pi leq theta leq 2pi) = frac{|pi – arccos(w)|}{pi} \$

However, I’m not sure how I’m supposed to proceed. I don’t see how this relates, since it doesn’t introduce \$mathop{dx}\$? Could someone please help me from here?

All topic