$ n$-dimensional rotation matrix I want to find $n$ dimensional rotation matrix which corresponds rotation of an angle  $\theta$
around the 
$(n−2)$-dimensional subspace.
There is  the n-dimensional rotation matrix formula. (see equation $15$)

$$I+(n_2n_1^T-n_1n_2^T)\sin(a)+(n_1n_1^T+n_2n_2^T)(\cos(a)-1)$$

where $n_1$ and $n_2$  are $n$-dimensional orthogonal unit vectors. 
Can anybody explain how can I use this formula,  for $n=6$? 
 A: Here's an example application using Python / Numpy:
import numpy as np

# input vectors
v1 = np.array( [1,1,1,1,1,1] )
v2 = np.array( [2,3,4,5,6,7] )

# Gram-Schmidt orthogonalization
n1 = v1 / np.linalg.norm(v1)
v2 = v2 - np.dot(n1,v2) * n1
n2 = v2 / np.linalg.norm(v2)

# rotation by pi/2
a = np.pi/2

I = np.identity(6)

R = I + ( np.outer(n2,n1) - np.outer(n1,n2) ) * np.sin(a) + ( np.outer(n1,n1) + np.outer(n2,n2) ) * (np.cos(a)-1)

# check result
print( np.matmul( R, n1 ) )
print( n2 )

See the result here.
A: Depends upon what you have at your disposal for calculating?
If you have a scientific programming language, like matlab, you construct an $n\times n$ dimensional matrix from the above formula. For example, I is just the identity, $n_2 n_1^T$ is an $n\times n$ dimensional matrix obtained by taking the product of a column vector with a row vector, etc... The resulting matrix performs an orthogonal rotation by angle $a$ in the plane spanned by $(n_1,n_2)$ and with orientation given by that couple.
