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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def _vector_close(vec1, vec2, eps=_EPS):
 r"""Checks that two vectors are equal to some threshold. Does so by computing :math:`s_1 = \|v_1\|_2` and :math:`s_2 = \|v_2\|_2` and then checking if .. math:: \|v_1 - v_2\|_2 \leq \varepsilon \min(s_1, s_2) where :math:`\varepsilon = 2^{-40} \approx 10^{-12}` is a fixed threshold. In the rare case that one of ``vec1`` or ``vec2`` is the zero vector (i.e. when :math:`\min(s_1, s_2) = 0`) instead checks that the other vector is close enough to zero: .. math:: \|v_1\|_2 = 0 \Longrightarrow \|v_2\|_2 \leq \varepsilon .. note:: This function assumes that both vectors have finite values, i.e. that no NaN or infinite numbers occur. NumPy provides :func:`np.allclose` for coverage of **all** cases. .. note:: There is also a Fortran implementation of this function, which will be used if it can be built. Args: vec1 (numpy.ndarray):
 First vector (1D) for comparison. vec2 (numpy.ndarray):
 Second vector (1D) for comparison. eps (float):
 Error threshold. Defaults to :math:`2^{-40}`. Returns: bool: Flag indicating if they are close to precision. """

    # NOTE: This relies on ``vec1`` and ``vec2`` being one-dimensional
    #       vectors so NumPy doesn't try to use a matrix norm.
    size1 = np.linalg.norm(vec1, ord=2)
    size2 = np.linalg.norm(vec2, ord=2)
    if size1 == 0:
        return size2 <= eps

    elif size2 == 0:
        return size1 <= eps

    else:
        upper_bound = eps * min(size1, size2)
        return np.linalg.norm(vec1 - vec2, ord=2) <= upper_bound