Transformation from Geocentric Coordinates into Equatorial Coordinates?
If I understand your question correctly than your problem is that the primary direction of vCR[3] is the intersection between the prime meridian and the Earth's equator, instead of the vernal equinox. The geographical latitude is the same as the declination, but to get the right ascension you will need to rotate your coordinates around the z-axis by an amount that is equal to the sidereal time at Greenwich ($\theta_G$). This is done, simply by replacing the the longitude (lon) with the sidereal time at your location $\theta = \theta_G-lon$.

x =  R* cos(lat) * cos(theta);
y=   R* cos(lat) * sin(theta);
z =  R* sin(lat);

vCR[3]={x,y,z};


where theta is the sidereal time $\theta$. 

The sidereal time is not a time but actually an angle that gives the right ascension of objects that are at that moment crossing the local meridian (i.e. the line between South and the Celestial north pole, for observers on the northern hemisphere). It is in a sense the longitude of your position on Earth if the geographical coordinate system were to rotate along with the celestial sphere.

EDIT: To calculate the Sidereal time, you can use the method explained in Astronomical Algorithms by Jean Meeus:

$$T = \frac{JD-2451545.0}{3625}$$
$$\theta_0 = 100.46061837 + 36000.770053608 T + 0.000387933 T^2 - \frac{T^3}{38710000}$$

The sidereal time is expressed in degrees.

Then express the UTC time as an angle: 

$$t = 15h+0.25m+\frac{0.25}{60}s$$

where $h$ is the number of hours, $m$, the number of minutes, and $s$ the number of seconds.

The sidereal time at Greenwich at the specified time is then:

$$\theta_G = \theta + 1.00273790935*t $$

and the local sidereal time is:

$$\theta = \theta_G - L$$

Where $L$ is the geographical longitude (positive west from Greenwich).